GSEB Class 12 Maths Solutions Chapter 7 Integrals Exercise 7.4

Integrate the following functions:

Question 1. \( \frac{3 x^{2}}{x^{6}+1} \)
Answer: Let \( I = \int \frac{3x^2}{x^6+1} \, dx \).
Put \( x^3 = t \), so that \( 3x^2 \, dx = dt \).
\( I = \int \frac{dt}{t^2+1} \)
\( \implies I = \tan^{-1}t + C \)
\( \implies I = \tan^{-1}x^3 + C \).
In simple words: We substitute \( x^3 \) with a new variable \( t \). Then we find the derivative \( dt \) for \( dx \). After this, the integral becomes a standard form that we can solve using the formula for \( \tan^{-1} \). Finally, we put \( x^3 \) back in place of \( t \).

Exam Tip: Recognise forms like \( \frac{f'(x)}{[f(x)]^2+a^2} \) to simplify integration by substitution. Always replace the substitute back at the end.

Question 2. \( \frac{1}{\sqrt{1+4 x^{2}}} \)
Answer: Let \( I = \int \frac{1}{\sqrt{1+4x^2}} \, dx \).
\( \implies I = \int \frac{1}{\sqrt{1+(2x)^2}} \, dx \)
Put \( 2x = t \), so that \( 2 \, dx = dt \), which means \( dx = \frac{dt}{2} \).
\( I = \int \frac{1}{\sqrt{1+t^2}} \frac{dt}{2} \)
\( \implies I = \frac{1}{2} \int \frac{1}{\sqrt{t^2+1^2}} \, dt \)
Using the formula \( \int \frac{1}{\sqrt{x^2+a^2}} \, dx = \log|x+\sqrt{x^2+a^2}| + C \):
\( \implies I = \frac{1}{2} \log|t+\sqrt{t^2+1}| + C \)
Substituting \( t = 2x \) back:
\( \implies I = \frac{1}{2} \log|2x+\sqrt{(2x)^2+1}| + C \)
\( \implies I = \frac{1}{2} \log|2x+\sqrt{4x^2+1}| + C \).

2nd Method:
Let \( I = \int \frac{1}{\sqrt{1+4x^2}} \, dx = \int \frac{dx}{\sqrt{1+(2x)^2}} \).
Put \( 2x = \tan\theta \) so that \( 2 \, dx = \sec^2\theta \, d\theta \).
\( \implies dx = \frac{\sec^2\theta}{2} \, d\theta \).
\( I = \int \frac{1}{\sqrt{1+\tan^2\theta}} \frac{\sec^2\theta}{2} \, d\theta \)
\( \implies I = \frac{1}{2} \int \frac{\sec^2\theta}{\sqrt{\sec^2\theta}} \, d\theta \)
\( \implies I = \frac{1}{2} \int \frac{\sec^2\theta}{\sec\theta} \, d\theta \)
\( \implies I = \frac{1}{2} \int \sec\theta \, d\theta \)
\( \implies I = \frac{1}{2} \log|\sec\theta + \tan\theta| + C \).
From \( 2x = \tan\theta \), we have \( \sec\theta = \sqrt{1+\tan^2\theta} = \sqrt{1+(2x)^2} = \sqrt{1+4x^2} \).
Substituting back:
\( \implies I = \frac{1}{2} \log| \sqrt{1+4x^2} + 2x | + C \).
In simple words: First, we write \( 4x^2 \) as \( (2x)^2 \). Then we use substitution by setting \( 2x \) to \( t \), which helps simplify the integral. We then use a standard integration formula involving \( \log \) and replace \( t \) back with \( 2x \). An alternate method uses trigonometric substitution, letting \( 2x = \tan\theta \), which also gives the same answer after converting back to \( x \).

Exam Tip: Be familiar with both algebraic substitution and trigonometric substitution methods for square root integrals, as one might be simpler depending on the form. Remember \( \int \frac{dx}{\sqrt{a^2+x^2}} = \log|x+\sqrt{a^2+x^2}| \).

Question 3. \( \frac{dx}{\sqrt{(2-x)^{2}+1}} \)
Answer: Let \( I = \int \frac{dx}{\sqrt{(2-x)^2+1}} \).
Put \( 2-x = t \).
Then \( -dx = dt \), so \( dx = -dt \).
\( I = \int \frac{-dt}{\sqrt{t^2+1}} \)
\( \implies I = - \int \frac{dt}{\sqrt{t^2+1^2}} \)
Using the formula \( \int \frac{1}{\sqrt{x^2+a^2}} \, dx = \log|x+\sqrt{x^2+a^2}| + C \):
\( \implies I = - \log|t+\sqrt{t^2+1}| + C \)
Substitute \( t = 2-x \) back:
\( \implies I = - \log|(2-x)+\sqrt{(2-x)^2+1}| + C \).
This can also be written as:
\( \implies I = \log \left| \frac{1}{(2-x)+\sqrt{(2-x)^2+1}} \right| + C \).
2nd Method:
Let \( 2-x = \tan\theta \). Then \( -dx = \sec^2\theta \, d\theta \), so \( dx = -\sec^2\theta \, d\theta \).
\( I = \int \frac{-\sec^2\theta \, d\theta}{\sqrt{\tan^2\theta+1}} \)
\( \implies I = \int \frac{-\sec^2\theta \, d\theta}{\sec\theta} \)
\( \implies I = - \int \sec\theta \, d\theta \)
\( \implies I = - \log|\sec\theta+\tan\theta| + C \).
From \( 2-x = \tan\theta \), we have \( \sec\theta = \sqrt{1+\tan^2\theta} = \sqrt{1+(2-x)^2} \).
Substituting back:
\( \implies I = - \log|\sqrt{1+(2-x)^2} + (2-x)| + C \).
In simple words: We replace \( 2-x \) with \( t \), which changes \( dx \) to \( -dt \). The integral then transforms into a known form involving \( \log \). After solving, we put \( 2-x \) back in place of \( t \) to get the final answer. An alternative way using trigonometric substitution also provides the same result.

Exam Tip: When \( dx \) becomes \( -dt \), do not forget the negative sign outside the integral. Always simplify the expression under the square root before applying the formula.

Question 4. \( \frac{dx}{\sqrt{9-25 x^{2}}} \)
Answer: Let \( I = \int \frac{dx}{\sqrt{9-25x^2}} \).
Rewrite the denominator: \( \sqrt{9-25x^2} = \sqrt{9 - (5x)^2} \).
Put \( 5x = t \).
Then \( 5 \, dx = dt \), so \( dx = \frac{dt}{5} \).
\( I = \int \frac{dt/5}{\sqrt{9-t^2}} \)
\( \implies I = \frac{1}{5} \int \frac{dt}{\sqrt{3^2-t^2}} \)
Using the formula \( \int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + C \):
\( \implies I = \frac{1}{5} \sin^{-1}\left(\frac{t}{3}\right) + C \)
Substitute \( t = 5x \) back:
\( \implies I = \frac{1}{5} \sin^{-1}\left(\frac{5x}{3}\right) + C \).
In simple words: We rewrite the expression to identify the form \( \sqrt{a^2-x^2} \). By substituting \( 5x \) with \( t \), the integral becomes a simple form for \( \sin^{-1} \). After solving, we replace \( t \) with \( 5x \) to get the final answer.

Exam Tip: For integrals of the form \( \frac{dx}{\sqrt{a^2-(bx)^2}} \), always perform the substitution \( bx=t \) first to get it into the standard form \( \frac{1}{b}\int \frac{dt}{\sqrt{a^2-t^2}} \).

Question 5. \( \frac{3 x}{1+2x^{4}} \)
Answer: Let \( I = \int \frac{3x}{1+2x^4} \, dx \).
Rewrite \( 2x^4 \) as \( 2(x^2)^2 \).
Put \( x^2 = t \).
Then \( 2x \, dx = dt \), so \( x \, dx = \frac{dt}{2} \).
\( I = \int \frac{3}{1+2t^2} \left(\frac{dt}{2}\right) \)
\( \implies I = \frac{3}{2} \int \frac{dt}{1+2t^2} \)
Factor out 2 from the denominator:
\( \implies I = \frac{3}{2} \int \frac{dt}{2\left(\frac{1}{2}+t^2\right)} \)
\( \implies I = \frac{3}{4} \int \frac{dt}{\left(\frac{1}{\sqrt{2}}\right)^2+t^2} \)
Using the formula \( \int \frac{1}{a^2+x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \):
\( \implies I = \frac{3}{4} \left[ \frac{1}{\frac{1}{\sqrt{2}}} \tan^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right) \right] + C \)
\( \implies I = \frac{3}{4} \left[ \sqrt{2} \tan^{-1}(\sqrt{2}t) \right] + C \)
\( \implies I = \frac{3\sqrt{2}}{4} \tan^{-1}(\sqrt{2}t) + C \)
Substitute \( t = x^2 \) back:
\( \implies I = \frac{3\sqrt{2}}{4} \tan^{-1}(\sqrt{2}x^2) + C \).
In simple words: We recognize that \( x^4 \) can be written as \( (x^2)^2 \). By substituting \( x^2 \) with \( t \), the integral simplifies to a form solvable using the \( \tan^{-1} \) formula. We then substitute \( t \) back with \( x^2 \) to get the final answer.

Exam Tip: For integrals involving \( x^n \) in the denominator and \( x^{n/2-1} \) in the numerator (like \( x^4 \) and \( x \)), consider substituting \( x^{n/2} = t \). Remember to simplify the constant term in the denominator to \( a^2 \) form if not already a perfect square.

Question 6. \( \frac{x^{2}}{1-x^{6}} \)
Answer: Let \( I = \int \frac{x^2}{1-x^6} \, dx \).
Rewrite \( x^6 \) as \( (x^3)^2 \).
Put \( x^3 = t \).
Then \( 3x^2 \, dx = dt \), so \( x^2 \, dx = \frac{dt}{3} \).
\( I = \int \frac{1}{1-t^2} \left(\frac{dt}{3}\right) \)
\( \implies I = \frac{1}{3} \int \frac{dt}{1^2-t^2} \)
Using the formula \( \int \frac{1}{a^2-x^2} \, dx = \frac{1}{2a} \log\left|\frac{a+x}{a-x}\right| + C \):
\( \implies I = \frac{1}{3} \left[ \frac{1}{2(1)} \log\left|\frac{1+t}{1-t}\right| \right] + C \)
\( \implies I = \frac{1}{6} \log\left|\frac{1+t}{1-t}\right| + C \)
Substitute \( t = x^3 \) back:
\( \implies I = \frac{1}{6} \log\left|\frac{1+x^3}{1-x^3}\right| + C \).
In simple words: We change \( x^6 \) to \( (x^3)^2 \). Then we replace \( x^3 \) with \( t \), which makes the integral a standard form that uses the \( \log \) function. After integrating, we swap \( t \) back with \( x^3 \) to get the final result.

Exam Tip: Look for opportunities to write terms like \( x^6 \) as squares or cubes of simpler terms (e.g., \( (x^3)^2 \)) to prepare for substitution. Remember the formula for \( \int \frac{dx}{a^2-x^2} \).

Question 7. \( \frac{x-1}{\sqrt{x^{2}-1}} \)
Answer: Let \( I = \int \frac{x-1}{\sqrt{x^2-1}} \, dx \).
Separate the integral into two parts:
\( I = \int \frac{x}{\sqrt{x^2-1}} \, dx - \int \frac{1}{\sqrt{x^2-1}} \, dx \)
Let \( I = I_1 - I_2 \).
For \( I_1 = \int \frac{x}{\sqrt{x^2-1}} \, dx \):
Put \( x^2-1 = t \).
Then \( 2x \, dx = dt \), so \( x \, dx = \frac{dt}{2} \).
\( I_1 = \int \frac{1}{\sqrt{t}} \left(\frac{dt}{2}\right) \)
\( \implies I_1 = \frac{1}{2} \int t^{-1/2} \, dt \)
\( \implies I_1 = \frac{1}{2} \left[ \frac{t^{1/2}}{1/2} \right] + C_1 \)
\( \implies I_1 = \sqrt{t} + C_1 \)
Substitute \( t = x^2-1 \) back:
\( \implies I_1 = \sqrt{x^2-1} + C_1 \).
For \( I_2 = \int \frac{1}{\sqrt{x^2-1}} \, dx \):
Using the formula \( \int \frac{1}{\sqrt{x^2-a^2}} \, dx = \log|x+\sqrt{x^2-a^2}| + C \):
\( \implies I_2 = \log|x+\sqrt{x^2-1}| + C_2 \).
Now, combine \( I_1 \) and \( I_2 \):
\( I = I_1 - I_2 = \sqrt{x^2-1} - \log|x+\sqrt{x^2-1}| + C \), where \( C = C_1 - C_2 \).
In simple words: We split the integral into two simpler parts. For the first part, we use a substitution to solve it. For the second part, we use a direct formula involving \( \log \). Finally, we combine these two results to obtain the complete answer.

Exam Tip: When the numerator is a sum or difference, split the fraction into separate integrals. This allows applying different techniques or formulas to each part, simplifying the process.

Question 8. \( \frac{x^{2}}{\sqrt{x^{6}+a^{6}}} \)
Answer: Let \( I = \int \frac{x^2}{\sqrt{x^6+a^6}} \, dx \).
Rewrite \( x^6 \) as \( (x^3)^2 \) and \( a^6 \) as \( (a^3)^2 \).
\( \implies I = \int \frac{x^2}{\sqrt{(x^3)^2+(a^3)^2}} \, dx \)
Put \( x^3 = t \).
Then \( 3x^2 \, dx = dt \), so \( x^2 \, dx = \frac{dt}{3} \).
\( I = \int \frac{1}{\sqrt{t^2+(a^3)^2}} \left(\frac{dt}{3}\right) \)
\( \implies I = \frac{1}{3} \int \frac{dt}{\sqrt{t^2+(a^3)^2}} \)
Using the formula \( \int \frac{1}{\sqrt{x^2+A^2}} \, dx = \log|x+\sqrt{x^2+A^2}| + C \), where \( A = a^3 \):
\( \implies I = \frac{1}{3} \log|t+\sqrt{t^2+(a^3)^2}| + C \)
Substitute \( t = x^3 \) back:
\( \implies I = \frac{1}{3} \log|x^3+\sqrt{(x^3)^2+(a^3)^2}| + C \)
\( \implies I = \frac{1}{3} \log|x^3+\sqrt{x^6+a^6}| + C \).
In simple words: We rewrite \( x^6 \) and \( a^6 \) as squares of \( x^3 \) and \( a^3 \) respectively. Then, we substitute \( x^3 \) with \( t \), which simplifies the integral to a standard form. We then use the \( \log \) formula and finally put \( x^3 \) back in place of \( t \) to get the answer.

Exam Tip: Similar to Question 5, look for powers that can be written as squares of a substitution variable. The presence of \( x^2 \) in the numerator suggests a substitution involving \( x^3 \).

Question 9. \( \frac{1}{\sqrt{7-6 x-x^{2}}} \)
Answer: Let \( I = \int \frac{1}{\sqrt{7-6x-x^2}} \, dx \).
To solve this, complete the square in the denominator:
\( 7-6x-x^2 = 7 - (x^2+6x) \)
\( = 7 - (x^2+6x+9-9) \)
\( = 7 - ( (x+3)^2 - 9 ) \)
\( = 7 - (x+3)^2 + 9 \)
\( = 16 - (x+3)^2 \)
So, \( I = \int \frac{dx}{\sqrt{16-(x+3)^2}} \)
\( \implies I = \int \frac{dx}{\sqrt{4^2-(x+3)^2}} \)
Using the formula \( \int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + C \):
Here, \( a=4 \) and the variable part is \( (x+3) \).
\( \implies I = \sin^{-1}\left(\frac{x+3}{4}\right) + C \).
In simple words: First, we complete the square for the expression under the square root in the denominator. This helps transform the integral into a standard form. Once in the standard form, we apply the formula for \( \sin^{-1} \) to find the answer.

Exam Tip: For quadratic expressions under a square root in the denominator, always complete the square to convert it into one of the standard integral forms like \( \sqrt{a^2-x^2} \), \( \sqrt{x^2-a^2} \), or \( \sqrt{x^2+a^2} \).

Question 10. \( \frac{1}{\sqrt{x^{2}+2x+2}} \)
Answer: Let \( I = \int \frac{1}{\sqrt{x^2+2x+2}} \, dx \).
Complete the square in the denominator:
\( x^2+2x+2 = (x^2+2x+1)+1 \)
\( = (x+1)^2+1 \)
So, \( I = \int \frac{dx}{\sqrt{(x+1)^2+1^2}} \)
Using the formula \( \int \frac{1}{\sqrt{x^2+a^2}} \, dx = \log|x+\sqrt{x^2+a^2}| + C \):
Here, the variable part is \( (x+1) \) and \( a=1 \).
\( \implies I = \log|(x+1)+\sqrt{(x+1)^2+1}| + C \)
\( \implies I = \log|(x+1)+\sqrt{x^2+2x+1+1}| + C \)
\( \implies I = \log|(x+1)+\sqrt{x^2+2x+2}| + C \).
In simple words: We complete the square for the expression under the square root. This changes the integral into a recognizable standard form. Then, we use the specific \( \log \) formula for this form to find the final integral.

Exam Tip: When completing the square for an expression like \( x^2+Bx+C \), remember to add and subtract \( (\frac{B}{2})^2 \). The standard integral forms are key here.

Question 11. \( \frac{1}{9 x^{2}+6x+5} \)
Answer: Let \( I = \int \frac{1}{9x^2+6x+5} \, dx \).
Factor out 9 from the denominator:
\( I = \frac{1}{9} \int \frac{dx}{x^2+\frac{6}{9}x+\frac{5}{9}} \)
\( \implies I = \frac{1}{9} \int \frac{dx}{x^2+\frac{2}{3}x+\frac{5}{9}} \)
Complete the square in the denominator:
\( x^2+\frac{2}{3}x+\frac{5}{9} = \left(x^2+\frac{2}{3}x+\left(\frac{1}{3}\right)^2\right) - \left(\frac{1}{3}\right)^2 + \frac{5}{9} \)
\( = \left(x+\frac{1}{3}\right)^2 - \frac{1}{9} + \frac{5}{9} \)
\( = \left(x+\frac{1}{3}\right)^2 + \frac{4}{9} \)
\( = \left(x+\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 \)
So, \( I = \frac{1}{9} \int \frac{dx}{\left(x+\frac{1}{3}\right)^2+\left(\frac{2}{3}\right)^2} \)
Using the formula \( \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \):
Here, \( x \) is \( \left(x+\frac{1}{3}\right) \) and \( a = \frac{2}{3} \).
\( \implies I = \frac{1}{9} \left[ \frac{1}{\frac{2}{3}} \tan^{-1}\left(\frac{x+\frac{1}{3}}{\frac{2}{3}}\right) \right] + C \)
\( \implies I = \frac{1}{9} \left[ \frac{3}{2} \tan^{-1}\left(\frac{\frac{3x+1}{3}}{\frac{2}{3}}\right) \right] + C \)
\( \implies I = \frac{1}{9} \left[ \frac{3}{2} \tan^{-1}\left(\frac{3x+1}{2}\right) \right] + C \)
\( \implies I = \frac{1}{6} \tan^{-1}\left(\frac{3x+1}{2}\right) + C \).
In simple words: First, we factor out the coefficient of \( x^2 \) from the denominator. Then, we complete the square for the quadratic expression. This converts the integral into a standard form that can be solved using the \( \tan^{-1} \) formula.

Exam Tip: Always make the coefficient of \( x^2 \) unity (1) before completing the square. Be careful with fractions when completing the square and simplifying the \( \tan^{-1} \) argument.

Question 12. \( \frac{1}{\sqrt{7-6 x-x^{2}}} \)
Answer: Let \( I = \int \frac{dx}{\sqrt{7-6x-x^2}} \).
Complete the square for the expression inside the square root:
\( 7-6x-x^2 = 7 - (x^2+6x) \)
\( = 7 - (x^2+6x+9-9) \)
\( = 7 - ((x+3)^2-9) \)
\( = 7 - (x+3)^2 + 9 \)
\( = 16 - (x+3)^2 \)
\( = 4^2 - (x+3)^2 \)
So, \( I = \int \frac{dx}{\sqrt{4^2-(x+3)^2}} \)
Using the formula \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C \):
Here, \( a=4 \) and the variable is \( (x+3) \).
\( \implies I = \sin^{-1}\left(\frac{x+3}{4}\right) + C \).
In simple words: We rearrange the terms and complete the square in the denominator. This changes the integral into a standard form. Then, we apply the inverse sine formula to solve it.

Exam Tip: Completing the square is crucial when integrating functions with quadratic expressions under a square root. Remember to factor out the negative sign if \( x^2 \) has a negative coefficient, then complete the square for the positive quadratic term.

Question 13. \( \frac{1}{\sqrt{(x-1)(x-2)}} \)
Answer: Let \( I = \int \frac{dx}{\sqrt{(x-1)(x-2)}} \).
Expand the product in the denominator:
\( (x-1)(x-2) = x^2-2x-x+2 = x^2-3x+2 \).
So, \( I = \int \frac{dx}{\sqrt{x^2-3x+2}} \).
Complete the square for \( x^2-3x+2 \):
\( x^2-3x+2 = \left(x^2-3x+\left(\frac{3}{2}\right)^2\right) - \left(\frac{3}{2}\right)^2 + 2 \)
\( = \left(x-\frac{3}{2}\right)^2 - \frac{9}{4} + 2 \)
\( = \left(x-\frac{3}{2}\right)^2 - \frac{9}{4} + \frac{8}{4} \)
\( = \left(x-\frac{3}{2}\right)^2 - \frac{1}{4} \)
\( = \left(x-\frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \)
So, \( I = \int \frac{dx}{\sqrt{\left(x-\frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} \)
Using the formula \( \int \frac{dx}{\sqrt{x^2-a^2}} = \log|x+\sqrt{x^2-a^2}| + C \):
Here, \( x \) is \( \left(x-\frac{3}{2}\right) \) and \( a = \frac{1}{2} \).
\( \implies I = \log\left|\left(x-\frac{3}{2}\right)+\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right| + C \)
\( \implies I = \log\left|\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+2}\right| + C \).
In simple words: First, we multiply the terms in the denominator. Then, we complete the square for the resulting quadratic expression. This helps change the integral into a standard form. Finally, we use the specific \( \log \) formula for that form to get the solution.

Exam Tip: When given a product of linear factors under a square root, expand them first. Then, proceed with completing the square as usual. Ensure correct sign usage, especially with negative terms.

Question 14. \( \frac{1}{\sqrt{8+3 x-x^{2}}} \)
Answer: Let \( I = \int \frac{dx}{\sqrt{8+3x-x^2}} \).
Complete the square for the expression inside the square root:
\( 8+3x-x^2 = 8 - (x^2-3x) \)
\( = 8 - \left(x^2-3x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\right) \)
\( = 8 - \left(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right) \)
\( = 8 - \left(x-\frac{3}{2}\right)^2 + \frac{9}{4} \)
\( = \frac{32}{4} + \frac{9}{4} - \left(x-\frac{3}{2}\right)^2 \)
\( = \frac{41}{4} - \left(x-\frac{3}{2}\right)^2 \)
\( = \left(\frac{\sqrt{41}}{2}\right)^2 - \left(x-\frac{3}{2}\right)^2 \)
So, \( I = \int \frac{dx}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2 - \left(x-\frac{3}{2}\right)^2}} \)
Using the formula \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C \):
Here, \( a=\frac{\sqrt{41}}{2} \) and the variable is \( \left(x-\frac{3}{2}\right) \).
\( \implies I = \sin^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right) + C \)
\( \implies I = \sin^{-1}\left(\frac{\frac{2x-3}{2}}{\frac{\sqrt{41}}{2}}\right) + C \)
\( \implies I = \sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right) + C \).
In simple words: We rearrange the terms and complete the square for the quadratic expression under the square root. This process converts the integral into a standard form involving \( a^2-x^2 \). Then, we apply the inverse sine formula to find the answer.

Exam Tip: Be very careful when handling the negative sign before the \( x^2 \) term while completing the square. Always ensure the constant term \( a^2 \) is positive for the \( \sin^{-1} \) form.

Question 15. \( \frac{1}{\sqrt{(x-a)(x-b)}} \)
Answer: Let \( I = \int \frac{dx}{\sqrt{(x-a)(x-b)}} \).
Expand the product in the denominator:
\( (x-a)(x-b) = x^2-bx-ax+ab = x^2-(a+b)x+ab \).
So, \( I = \int \frac{dx}{\sqrt{x^2-(a+b)x+ab}} \).
Complete the square for \( x^2-(a+b)x+ab \):
\( x^2-(a+b)x+ab = \left(x^2-(a+b)x+\left(\frac{a+b}{2}\right)^2\right) - \left(\frac{a+b}{2}\right)^2 + ab \)
\( = \left(x-\frac{a+b}{2}\right)^2 - \frac{a^2+2ab+b^2}{4} + \frac{4ab}{4} \)
\( = \left(x-\frac{a+b}{2}\right)^2 - \frac{a^2-2ab+b^2}{4} \)
\( = \left(x-\frac{a+b}{2}\right)^2 - \frac{(a-b)^2}{4} \)
\( = \left(x-\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2 \)
So, \( I = \int \frac{dx}{\sqrt{\left(x-\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2}} \)
Using the formula \( \int \frac{dx}{\sqrt{x^2-A^2}} = \log|x+\sqrt{x^2-A^2}| + C \):
Here, \( x \) is \( \left(x-\frac{a+b}{2}\right) \) and \( A = \left(\frac{a-b}{2}\right) \).
\( \implies I = \log\left|\left(x-\frac{a+b}{2}\right)+\sqrt{\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2}\right| + C \)
\( \implies I = \log\left|\left(x-\frac{a+b}{2}\right)+\sqrt{(x-a)(x-b)}\right| + C \).
In simple words: First, we expand the product in the denominator. Then, we complete the square for the resulting quadratic expression, which involves terms with \( a \) and \( b \). This transforms the integral into a standard \( \log \) form. Finally, we apply the formula and simplify to get the answer.

Exam Tip: Be careful with algebraic manipulation when completing the square with literal coefficients (a and b). Keep track of signs and denominators, especially when combining fractions.

Question 16. \( \frac{4 x+1}{\sqrt{2 x^{2}+x-3}} \)
Answer: Let \( I = \int \frac{4x+1}{\sqrt{2x^2+x-3}} \, dx \).
We observe that the derivative of \( 2x^2+x-3 \) is \( 4x+1 \).
Let \( t = 2x^2+x-3 \).
Then \( dt = (4x+1) \, dx \).
The integral becomes:
\( I = \int \frac{dt}{\sqrt{t}} \)
\( \implies I = \int t^{-1/2} \, dt \)
\( \implies I = \frac{t^{1/2}}{1/2} + C \)
\( \implies I = 2\sqrt{t} + C \)
Substitute \( t = 2x^2+x-3 \) back:
\( \implies I = 2\sqrt{2x^2+x-3} + C \).
In simple words: We notice that the numerator is exactly the derivative of the expression inside the square root in the denominator. This allows us to use a direct substitution, letting the expression under the root be \( t \). The integral then simplifies to a basic power rule, giving the answer quickly.

Exam Tip: Always check if the numerator is the derivative (or a constant multiple of the derivative) of the expression in the denominator or under a radical. This is a common and efficient substitution method.

Question 17. \( \frac{x+2}{\sqrt{x^{2}-1}} \)
Answer: Let \( I = \int \frac{x+2}{\sqrt{x^2-1}} \, dx \).
Separate the integral into two parts:
\( I = \int \frac{x}{\sqrt{x^2-1}} \, dx + \int \frac{2}{\sqrt{x^2-1}} \, dx \)
Let \( I = I_1 + I_2 \).
For \( I_1 = \int \frac{x}{\sqrt{x^2-1}} \, dx \):
Put \( x^2-1 = t \).
Then \( 2x \, dx = dt \), so \( x \, dx = \frac{dt}{2} \).
\( I_1 = \int \frac{1}{\sqrt{t}} \left(\frac{dt}{2}\right) \)
\( \implies I_1 = \frac{1}{2} \int t^{-1/2} \, dt \)
\( \implies I_1 = \frac{1}{2} \left[ \frac{t^{1/2}}{1/2} \right] + C_1 \)
\( \implies I_1 = \sqrt{t} + C_1 \)
Substitute \( t = x^2-1 \) back:
\( \implies I_1 = \sqrt{x^2-1} + C_1 \).
For \( I_2 = \int \frac{2}{\sqrt{x^2-1}} \, dx \):
\( \implies I_2 = 2 \int \frac{1}{\sqrt{x^2-1^2}} \, dx \)
Using the formula \( \int \frac{1}{\sqrt{x^2-a^2}} \, dx = \log|x+\sqrt{x^2-a^2}| + C \):
\( \implies I_2 = 2 \log|x+\sqrt{x^2-1}| + C_2 \).
Now, combine \( I_1 \) and \( I_2 \):
\( I = I_1 + I_2 = \sqrt{x^2-1} + 2 \log|x+\sqrt{x^2-1}| + C \), where \( C = C_1 + C_2 \).
In simple words: We split the integral into two parts. The first part is solved using a simple substitution, while the second part uses a direct formula. Combining the results of both parts gives the final answer for the integral.

Exam Tip: Splitting the integral is effective when the numerator is a linear expression and the denominator is a quadratic under a square root. This often yields one integral solvable by substitution and another by a standard formula.

Question 18. \( \frac{5 x-2}{1+2 x+3 x^{2}} \)
Answer: Let \( I = \int \frac{5x-2}{3x^2+2x+1} \, dx \).
Let \( N = 5x-2 \) and \( D = 3x^2+2x+1 \).
We need to express \( 5x-2 \) as \( A \frac{d}{dx}(3x^2+2x+1) + B \).
\( 5x-2 = A(6x+2) + B \)
\( 5x-2 = 6Ax + 2A + B \)
Comparing coefficients of \( x \): \( 6A = 5 \implies A = \frac{5}{6} \).
Comparing constant terms: \( 2A+B = -2 \).
Substitute \( A=\frac{5}{6} \): \( 2\left(\frac{5}{6}\right)+B = -2 \).
\( \frac{10}{6}+B = -2 \implies \frac{5}{3}+B = -2 \).
\( B = -2 - \frac{5}{3} = -\frac{6}{3} - \frac{5}{3} = -\frac{11}{3} \).
So, \( I = \int \frac{\frac{5}{6}(6x+2) - \frac{11}{3}}{3x^2+2x+1} \, dx \)
\( \implies I = \frac{5}{6} \int \frac{6x+2}{3x^2+2x+1} \, dx - \frac{11}{3} \int \frac{1}{3x^2+2x+1} \, dx \)
Let \( I = I_1 - I_2 \).
For \( I_1 = \frac{5}{6} \int \frac{6x+2}{3x^2+2x+1} \, dx \):
Let \( u = 3x^2+2x+1 \). Then \( du = (6x+2) \, dx \).
\( I_1 = \frac{5}{6} \int \frac{du}{u} = \frac{5}{6} \log|u| + C_1 \)
\( \implies I_1 = \frac{5}{6} \log|3x^2+2x+1| + C_1 \).
For \( I_2 = \frac{11}{3} \int \frac{1}{3x^2+2x+1} \, dx \):
Factor out 3 from the denominator:
\( I_2 = \frac{11}{3} \int \frac{1}{3\left(x^2+\frac{2}{3}x+\frac{1}{3}\right)} \, dx \)
\( \implies I_2 = \frac{11}{9} \int \frac{1}{x^2+\frac{2}{3}x+\frac{1}{3}} \, dx \)
Complete the square for \( x^2+\frac{2}{3}x+\frac{1}{3} \):
\( x^2+\frac{2}{3}x+\frac{1}{3} = \left(x^2+\frac{2}{3}x+\left(\frac{1}{3}\right)^2\right) - \left(\frac{1}{3}\right)^2 + \frac{1}{3} \)
\( = \left(x+\frac{1}{3}\right)^2 - \frac{1}{9} + \frac{1}{3} \)
\( = \left(x+\frac{1}{3}\right)^2 - \frac{1}{9} + \frac{3}{9} \)
\( = \left(x+\frac{1}{3}\right)^2 + \frac{2}{9} \)
\( = \left(x+\frac{1}{3}\right)^2 + \left(\frac{\sqrt{2}}{3}\right)^2 \)
So, \( I_2 = \frac{11}{9} \int \frac{1}{\left(x+\frac{1}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2} \, dx \)
Using the formula \( \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \):
Here, \( x \) is \( \left(x+\frac{1}{3}\right) \) and \( a = \frac{\sqrt{2}}{3} \).
\( I_2 = \frac{11}{9} \left[ \frac{1}{\frac{\sqrt{2}}{3}} \tan^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right) \right] + C_2 \)
\( \implies I_2 = \frac{11}{9} \left[ \frac{3}{\sqrt{2}} \tan^{-1}\left(\frac{\frac{3x+1}{3}}{\frac{\sqrt{2}}{3}}\right) \right] + C_2 \)
\( \implies I_2 = \frac{11}{3\sqrt{2}} \tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) + C_2 \).
Combine \( I_1 \) and \( I_2 \):
\( I = \frac{5}{6} \log|3x^2+2x+1| - \frac{11}{3\sqrt{2}} \tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) + C \), where \( C = C_1-C_2 \).
In simple words: We express the numerator in terms of the derivative of the denominator, plus a constant. This splits the integral into two parts. The first part is solved by direct substitution, and the second part requires completing the square in the denominator and then using the \( \tan^{-1} \) formula. Finally, we combine these results to get the total integral.

Exam Tip: For integrals of the form \( \int \frac{Px+Q}{Ax^2+Bx+C} \, dx \), always use the method \( Px+Q = A'(2Ax+B) + B' \). This breaks the integral into two standard forms: one logarithmic and one arctangent.

Question 19. \( \frac{6 x+7}{\sqrt{(x-5)(x-4)}} \)
Answer: Let \( I = \int \frac{6x+7}{\sqrt{(x-5)(x-4)}} \, dx \).
First, expand the product in the denominator:
\( (x-5)(x-4) = x^2-4x-5x+20 = x^2-9x+20 \).
So, \( I = \int \frac{6x+7}{\sqrt{x^2-9x+20}} \, dx \).
Let \( N = 6x+7 \) and \( D = x^2-9x+20 \).
We need to express \( 6x+7 \) as \( A \frac{d}{dx}(x^2-9x+20) + B \).
\( 6x+7 = A(2x-9) + B \)
\( 6x+7 = 2Ax - 9A + B \)
Comparing coefficients of \( x \): \( 2A = 6 \implies A = 3 \).
Comparing constant terms: \( -9A+B = 7 \).
Substitute \( A=3 \): \( -9(3)+B = 7 \).
\( -27+B = 7 \implies B = 7+27 = 34 \).
So, \( I = \int \frac{3(2x-9) + 34}{\sqrt{x^2-9x+20}} \, dx \)
\( \implies I = 3 \int \frac{2x-9}{\sqrt{x^2-9x+20}} \, dx + 34 \int \frac{1}{\sqrt{x^2-9x+20}} \, dx \)
Let \( I = 3I_1 + 34I_2 \).
For \( I_1 = \int \frac{2x-9}{\sqrt{x^2-9x+20}} \, dx \):
Let \( u = x^2-9x+20 \). Then \( du = (2x-9) \, dx \).
\( I_1 = \int \frac{du}{\sqrt{u}} = \int u^{-1/2} \, du \)
\( \implies I_1 = \frac{u^{1/2}}{1/2} + C_1 = 2\sqrt{u} + C_1 \)
\( \implies I_1 = 2\sqrt{x^2-9x+20} + C_1 \).
For \( I_2 = \int \frac{1}{\sqrt{x^2-9x+20}} \, dx \):
Complete the square for \( x^2-9x+20 \):
\( x^2-9x+20 = \left(x^2-9x+\left(\frac{9}{2}\right)^2\right) - \left(\frac{9}{2}\right)^2 + 20 \)
\( = \left(x-\frac{9}{2}\right)^2 - \frac{81}{4} + \frac{80}{4} \)
\( = \left(x-\frac{9}{2}\right)^2 - \frac{1}{4} \)
\( = \left(x-\frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \)
So, \( I_2 = \int \frac{dx}{\sqrt{\left(x-\frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} \)
Using the formula \( \int \frac{dx}{\sqrt{x^2-a^2}} = \log|x+\sqrt{x^2-a^2}| + C \):
Here, \( x \) is \( \left(x-\frac{9}{2}\right) \) and \( a = \frac{1}{2} \).
\( \implies I_2 = \log\left|\left(x-\frac{9}{2}\right)+\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right| + C_2 \)
\( \implies I_2 = \log\left|\left(x-\frac{9}{2}\right)+\sqrt{x^2-9x+20}\right| + C_2 \).
Combine \( I_1 \) and \( I_2 \):
\( I = 3(2\sqrt{x^2-9x+20}) + 34 \log\left|\left(x-\frac{9}{2}\right)+\sqrt{x^2-9x+20}\right| + C \)
\( \implies I = 6\sqrt{x^2-9x+20} + 34 \log\left|\left(\frac{2x-9}{2}\right)+\sqrt{x^2-9x+20}\right| + C \), where \( C = 3C_1 + 34C_2 \).
In simple words: First, we expand the terms under the square root in the denominator. Then, we rewrite the numerator in a form that includes the derivative of the expression under the square root. This breaks the integral into two parts: one solved by simple substitution and another by completing the square and using the \( \log \) formula. Finally, we combine the results to get the complete answer.

Exam Tip: This type of integral (\( \int \frac{Px+Q}{\sqrt{Ax^2+Bx+C}} \, dx \)) is a standard form. Remember to express \( Px+Q \) as \( A'(2Ax+B)+B' \) to successfully split it into a substitution-based integral and a complete-the-square integral.

Question 15. Integrate the function \( \frac{1}{\sqrt{(x-a)(x-b)}} \)
Answer: Let the integral be \( I \).
\( I = \int \frac{dx}{\sqrt{(x-a)(x-b)}} \)
First, expand the term under the square root:
\( (x-a)(x-b) = x^2 - ax - bx + ab = x^2 - (a+b)x + ab \)
So, the integral becomes:
\( I = \int \frac{dx}{\sqrt{x^2 - (a+b)x + ab}} \)
Now, complete the square for the quadratic expression in the denominator:
\( x^2 - (a+b)x + ab = x^2 - 2 \left(\frac{a+b}{2}\right)x + \left(\frac{a+b}{2}\right)^2 - \left(\frac{a+b}{2}\right)^2 + ab \)
\( = \left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a+b}{2}\right)^2 + ab \)
\( = \left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a^2+2ab+b^2 - 4ab}{4}\right) \)
\( = \left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a^2-2ab+b^2}{4}\right) \)
\( = \left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2 \)
Let \( u = x - \frac{a+b}{2} \) and \( C_{val} = \frac{a-b}{2} \). Then \( du = dx \).
The integral becomes:
\( I = \int \frac{du}{\sqrt{u^2 - C_{val}^2}} \)
Using the standard integral formula \( \int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + \text{Constant} \):
\( I = \log \left| u + \sqrt{u^2 - C_{val}^2} \right| + \text{Constant} \)
Substitute back \( u = x - \frac{a+b}{2} \) and \( C_{val} = \frac{a-b}{2} \):
\( I = \log \left| \left(x - \frac{a+b}{2}\right) + \sqrt{\left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2} \right| + C \)
The term under the square root is the original quadratic expression:
\( \left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2 = x^2 - (a+b)x + ab = (x-a)(x-b) \)
So, the final solution is:
\( I = \log \left| x - \frac{a+b}{2} + \sqrt{(x-a)(x-b)} \right| + C \)
In simple words: To integrate this function, first multiply out the terms in the denominator. Then, use the method of completing the square to change the expression into a standard form. Once it's in a standard form, apply the appropriate integral formula for expressions involving square roots.

Exam Tip: Remember to complete the square carefully for quadratic expressions in the denominator. This technique helps transform the integrand into a standard form that matches common integration formulas. Always double-check your algebraic manipulations.

Question 20. Integrate the function \( \frac{x+2}{\sqrt{4x-x^2}} \)
Answer: Let the integral be \( I \).
\( I = \int \frac{x+2}{\sqrt{4x-x^2}} dx \)
Rewrite the numerator to match the derivative of the denominator's inner function or to separate into simpler integrals.
The derivative of \( 4x - x^2 \) is \( 4 - 2x \). We want to create \( (4-2x) \) in the numerator.
Let \( x+2 = A(4-2x) + B \)
\( x+2 = 4A - 2Ax + B \)
Comparing coefficients of \( x \): \( 1 = -2A \implies A = -\frac{1}{2} \)
Comparing constant terms: \( 2 = 4A + B \implies 2 = 4\left(-\frac{1}{2}\right) + B \implies 2 = -2 + B \implies B = 4 \)
So, \( x+2 = -\frac{1}{2}(4-2x) + 4 \)
Substitute this back into the integral:
\( I = \int \frac{-\frac{1}{2}(4-2x) + 4}{\sqrt{4x-x^2}} dx \)
\( I = -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}} dx + 4 \int \frac{1}{\sqrt{4x-x^2}} dx \)
Let \( I_1 = -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}} dx \) and \( I_2 = 4 \int \frac{1}{\sqrt{4x-x^2}} dx \).
For \( I_1 \): Let \( t = 4x-x^2 \), then \( dt = (4-2x) dx \).
\( I_1 = -\frac{1}{2} \int \frac{dt}{\sqrt{t}} = -\frac{1}{2} \int t^{-1/2} dt = -\frac{1}{2} \left( \frac{t^{1/2}}{1/2} \right) + C_1 = -\sqrt{t} + C_1 \)
Substitute back \( t = 4x-x^2 \):
\( I_1 = -\sqrt{4x-x^2} + C_1 \)
For \( I_2 \): Complete the square for \( 4x-x^2 \).
\( 4x-x^2 = -(x^2-4x) = -(x^2-4x+4-4) = -((x-2)^2-4) = 4-(x-2)^2 \)
So, \( I_2 = 4 \int \frac{1}{\sqrt{4-(x-2)^2}} dx \)
Using the standard integral formula \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1} \left(\frac{x}{a}\right) + C \):
Here, \( a=2 \) and \( x=(x-2) \).
\( I_2 = 4 \sin^{-1} \left(\frac{x-2}{2}\right) + C_2 \)
Combining \( I_1 \) and \( I_2 \):
\( I = I_1 + I_2 = -\sqrt{4x-x^2} + 4 \sin^{-1} \left(\frac{x-2}{2}\right) + C \), where \( C = C_1+C_2 \).
In simple words: First, adjust the top part of the fraction so it helps simplify the bottom part, separating the integral into two easier parts. For the first part, use substitution. For the second part, complete the square in the bottom expression to fit a known inverse sine integral formula. Then, put both results together.

Exam Tip: For integrals of the form \( \int \frac{px+q}{\sqrt{ax^2+bx+c}} dx \), always try to express \( px+q \) as \( A \cdot \frac{d}{dx}(ax^2+bx+c) + B \). This splits the integral into two standard forms: one solvable by substitution and the other by completing the square in the denominator.

Question 21. Integrate the function \( \frac{x+2}{\sqrt{x^2+2x+3}} \)
Answer: Let the integral be \( I \).
\( I = \int \frac{x+2}{\sqrt{x^2+2x+3}} dx \)
We express the numerator \( x+2 \) in the form \( A \cdot \frac{d}{dx}(x^2+2x+3) + B \).
The derivative of \( x^2+2x+3 \) is \( 2x+2 \).
So, \( x+2 = A(2x+2) + B \)
\( x+2 = 2Ax + 2A + B \)
Comparing coefficients of \( x \): \( 1 = 2A \implies A = \frac{1}{2} \)
Comparing constant terms: \( 2 = 2A + B \implies 2 = 2\left(\frac{1}{2}\right) + B \implies 2 = 1 + B \implies B = 1 \)
Thus, \( x+2 = \frac{1}{2}(2x+2) + 1 \)
Substitute this into the integral:
\( I = \int \frac{\frac{1}{2}(2x+2) + 1}{\sqrt{x^2+2x+3}} dx \)
\( I = \frac{1}{2} \int \frac{2x+2}{\sqrt{x^2+2x+3}} dx + \int \frac{1}{\sqrt{x^2+2x+3}} dx \)
Let \( I_1 = \frac{1}{2} \int \frac{2x+2}{\sqrt{x^2+2x+3}} dx \) and \( I_2 = \int \frac{1}{\sqrt{x^2+2x+3}} dx \).
For \( I_1 \): Let \( t = x^2+2x+3 \), then \( dt = (2x+2) dx \).
\( I_1 = \frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \left( \frac{t^{1/2}}{1/2} \right) + C_1 = \sqrt{t} + C_1 \)
Substitute back \( t = x^2+2x+3 \):
\( I_1 = \sqrt{x^2+2x+3} + C_1 \)
For \( I_2 \): Complete the square for \( x^2+2x+3 \).
\( x^2+2x+3 = x^2+2x+1+2 = (x+1)^2 + 2 = (x+1)^2 + (\sqrt{2})^2 \)
So, \( I_2 = \int \frac{1}{\sqrt{(x+1)^2 + (\sqrt{2})^2}} dx \)
Using the standard integral formula \( \int \frac{dx}{\sqrt{x^2+a^2}} = \log |x + \sqrt{x^2+a^2}| + C \):
Here, \( x = (x+1) \) and \( a = \sqrt{2} \).
\( I_2 = \log \left| (x+1) + \sqrt{(x+1)^2 + (\sqrt{2})^2} \right| + C_2 \)
Substitute back the simplified expression:
\( I_2 = \log \left| x+1 + \sqrt{x^2+2x+3} \right| + C_2 \)
Combining \( I_1 \) and \( I_2 \):
\( I = I_1 + I_2 = \sqrt{x^2+2x+3} + \log \left| x+1 + \sqrt{x^2+2x+3} \right| + C \), where \( C = C_1+C_2 \).
In simple words: Break the top part of the fraction into two terms: one that is the derivative of the expression inside the square root at the bottom, and another constant term. Integrate the first term using a simple substitution. For the second term, complete the square of the quadratic expression under the root to use a standard logarithmic integral formula. Combine these results for the final answer.

Exam Tip: When the denominator is \( \sqrt{ax^2+bx+c} \) and the numerator is a linear function, always use the technique of expressing the numerator as \( A \cdot (derivative \ of \ denominator) + B \). This allows you to split the integral into two solvable parts, one with direct substitution and the other with completing the square.

Question 22. Integrate the function \( \frac{x+3}{x^2-2x-5} \)
Answer: Let the integral be \( I \).
\( I = \int \frac{x+3}{x^2-2x-5} dx \)
We express the numerator \( x+3 \) in the form \( A \cdot \frac{d}{dx}(x^2-2x-5) + B \).
The derivative of \( x^2-2x-5 \) is \( 2x-2 \).
So, \( x+3 = A(2x-2) + B \)
\( x+3 = 2Ax - 2A + B \)
Comparing coefficients of \( x \): \( 1 = 2A \implies A = \frac{1}{2} \)
Comparing constant terms: \( 3 = -2A + B \implies 3 = -2\left(\frac{1}{2}\right) + B \implies 3 = -1 + B \implies B = 4 \)
Thus, \( x+3 = \frac{1}{2}(2x-2) + 4 \)
Substitute this into the integral:
\( I = \int \frac{\frac{1}{2}(2x-2) + 4}{x^2-2x-5} dx \)
\( I = \frac{1}{2} \int \frac{2x-2}{x^2-2x-5} dx + 4 \int \frac{1}{x^2-2x-5} dx \)
Let \( I_1 = \frac{1}{2} \int \frac{2x-2}{x^2-2x-5} dx \) and \( I_2 = 4 \int \frac{1}{x^2-2x-5} dx \).
For \( I_1 \): Let \( t = x^2-2x-5 \), then \( dt = (2x-2) dx \).
\( I_1 = \frac{1}{2} \int \frac{dt}{t} = \frac{1}{2} \log |t| + C_1 \)
Substitute back \( t = x^2-2x-5 \):
\( I_1 = \frac{1}{2} \log |x^2-2x-5| + C_1 \)
For \( I_2 \): Complete the square for \( x^2-2x-5 \).
\( x^2-2x-5 = x^2-2x+1-1-5 = (x-1)^2 - 6 = (x-1)^2 - (\sqrt{6})^2 \)
So, \( I_2 = 4 \int \frac{1}{(x-1)^2 - (\sqrt{6})^2} dx \)
Using the standard integral formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \):
Here, \( x=(x-1) \) and \( a=\sqrt{6} \).
\( I_2 = 4 \cdot \frac{1}{2\sqrt{6}} \log \left| \frac{(x-1)-\sqrt{6}}{(x-1)+\sqrt{6}} \right| + C_2 \)
\( I_2 = \frac{2}{\sqrt{6}} \log \left| \frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right| + C_2 \)
Combining \( I_1 \) and \( I_2 \):
\( I = I_1 + I_2 = \frac{1}{2} \log |x^2-2x-5| + \frac{2}{\sqrt{6}} \log \left| \frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right| + C \), where \( C = C_1+C_2 \).
In simple words: To integrate this rational function, first split the numerator into two parts: one being a multiple of the denominator's derivative, and the other a constant. The first integral uses a simple logarithmic substitution. For the second integral, complete the square in the denominator to use a standard logarithmic integral formula involving \( x^2 - a^2 \).

Exam Tip: For integrals of the form \( \int \frac{px+q}{ax^2+bx+c} dx \), always express the numerator as \( A \cdot \frac{d}{dx}(ax^2+bx+c) + B \). This transforms the integral into a sum of two standard forms: \( A \int \frac{f'(x)}{f(x)} dx \) (which gives a logarithm) and \( B \int \frac{1}{ax^2+bx+c} dx \) (which requires completing the square and using a standard formula).

Question 23. Integrate the function \( \frac{5x+3}{\sqrt{x^2+4x+10}} \)
Answer: Let the integral be \( I \).
\( I = \int \frac{5x+3}{\sqrt{x^2+4x+10}} dx \)
We express the numerator \( 5x+3 \) in the form \( A \cdot \frac{d}{dx}(x^2+4x+10) + B \).
The derivative of \( x^2+4x+10 \) is \( 2x+4 \).
So, \( 5x+3 = A(2x+4) + B \)
\( 5x+3 = 2Ax + 4A + B \)
Comparing coefficients of \( x \): \( 5 = 2A \implies A = \frac{5}{2} \)
Comparing constant terms: \( 3 = 4A + B \implies 3 = 4\left(\frac{5}{2}\right) + B \implies 3 = 10 + B \implies B = -7 \)
Thus, \( 5x+3 = \frac{5}{2}(2x+4) - 7 \)
Substitute this into the integral:
\( I = \int \frac{\frac{5}{2}(2x+4) - 7}{\sqrt{x^2+4x+10}} dx \)
\( I = \frac{5}{2} \int \frac{2x+4}{\sqrt{x^2+4x+10}} dx - 7 \int \frac{1}{\sqrt{x^2+4x+10}} dx \)
Let \( I_1 = \frac{5}{2} \int \frac{2x+4}{\sqrt{x^2+4x+10}} dx \) and \( I_2 = -7 \int \frac{1}{\sqrt{x^2+4x+10}} dx \).
For \( I_1 \): Let \( t = x^2+4x+10 \), then \( dt = (2x+4) dx \).
\( I_1 = \frac{5}{2} \int \frac{dt}{\sqrt{t}} = \frac{5}{2} \int t^{-1/2} dt = \frac{5}{2} \left( \frac{t^{1/2}}{1/2} \right) + C_1 = 5\sqrt{t} + C_1 \)
Substitute back \( t = x^2+4x+10 \):
\( I_1 = 5\sqrt{x^2+4x+10} + C_1 \)
For \( I_2 \): Complete the square for \( x^2+4x+10 \).
\( x^2+4x+10 = x^2+4x+4+6 = (x+2)^2 + 6 = (x+2)^2 + (\sqrt{6})^2 \)
So, \( I_2 = -7 \int \frac{1}{\sqrt{(x+2)^2 + (\sqrt{6})^2}} dx \)
Using the standard integral formula \( \int \frac{dx}{\sqrt{x^2+a^2}} = \log |x + \sqrt{x^2+a^2}| + C \):
Here, \( x=(x+2) \) and \( a=\sqrt{6} \).
\( I_2 = -7 \log \left| (x+2) + \sqrt{(x+2)^2 + (\sqrt{6})^2} \right| + C_2 \)
Substitute back the simplified expression:
\( I_2 = -7 \log \left| x+2 + \sqrt{x^2+4x+10} \right| + C_2 \)
Combining \( I_1 \) and \( I_2 \):
\( I = I_1 + I_2 = 5\sqrt{x^2+4x+10} - 7 \log \left| x+2 + \sqrt{x^2+4x+10} \right| + C \), where \( C = C_1+C_2 \).
In simple words: To integrate this expression, you first change the numerator so it includes the derivative of the term under the square root in the denominator, plus a constant. This splits the problem into two separate integrals. One integral can be solved using simple substitution, while the other requires completing the square in the denominator and then using a standard logarithmic integration formula.

Exam Tip: Always follow the systematic approach for integrals like \( \int \frac{px+q}{\sqrt{ax^2+bx+c}} dx \): adjust the numerator to be \( A \cdot (2ax+b) + B \). This ensures the integral is broken down into easily solvable forms: \( A \int \frac{f'(x)}{\sqrt{f(x)}} dx \) and \( B \int \frac{1}{\sqrt{ax^2+bx+c}} dx \). The second form always needs completing the square.

Question 24. Evaluate the integral \( \int\frac{dx}{x^2+2x+2} \). The result equals
(a) x tan¯¹(x + 1) + C
(b) tan¯¹(x + 1) + C
(c) (x + 1) tan¯¹x + C
(d) tan¯¹x + C
Answer: (b) tan¯¹(x + 1) + C
We can evaluate the integral by completing the square in the denominator:
\( \int \frac{dx}{x^2+2x+2} = \int \frac{dx}{(x^2+2x+1)+1} = \int \frac{dx}{(x+1)^2+1} \)
This is a standard integral of the form \( \int \frac{dx}{u^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C \).
Here, \( u = (x+1) \) and \( a = 1 \).
So, \( \int \frac{dx}{(x+1)^2+1} = \tan^{-1}(x+1) + C \).
Therefore, option (b) is the correct choice.
In simple words: Complete the square in the denominator to get \( (x+1)^2+1 \). This form directly integrates to \( \tan^{-1}(x+1) + C \), matching option (b).

Exam Tip: For integrals with a quadratic in the denominator and no square root, completing the square usually leads to an inverse tangent form. Recognize this pattern to quickly identify the solution strategy.

Question 25. Let \( I = \int\frac{d x}{\sqrt{9x-4x^2}} \). The result equals
(a) \( \frac{1}{9} \sin^{-1} \left(\frac{9x-8}{8}\right) + C \)
(b) \( \frac{1}{2} \sin^{-1} \left(\frac{8x-9}{9}\right) + C \)
(c) \( \frac{1}{3} \sin^{-1} \left(\frac{9x-8}{8}\right) + C \)
(d) \( \frac{1}{2} \sin^{-1} \left(\frac{9x-8}{8}\right) + C \)
Answer: (b) \( \frac{1}{2} \sin^{-1} \left(\frac{8x-9}{9}\right) + C \)
To evaluate this integral, we first simplify the quadratic expression under the square root by completing the square:
\( 9x-4x^2 = -4(x^2 - \frac{9}{4}x) \)
\( = -4 \left( x^2 - \frac{9}{4}x + \left(\frac{9}{8}\right)^2 - \left(\frac{9}{8}\right)^2 \right) \)
\( = -4 \left( \left(x - \frac{9}{8}\right)^2 - \frac{81}{64} \right) \)
\( = 4 \left( \frac{81}{64} - \left(x - \frac{9}{8}\right)^2 \right) \)
Now, take the square root:
\( \sqrt{9x-4x^2} = \sqrt{4 \left( \frac{81}{64} - \left(x - \frac{9}{8}\right)^2 \right)} = 2 \sqrt{\left(\frac{9}{8}\right)^2 - \left(x - \frac{9}{8}\right)^2} \)
Substitute this back into the integral:
\( I = \int \frac{dx}{2 \sqrt{\left(\frac{9}{8}\right)^2 - \left(x - \frac{9}{8}\right)^2}} = \frac{1}{2} \int \frac{dx}{\sqrt{\left(\frac{9}{8}\right)^2 - \left(x - \frac{9}{8}\right)^2}} \)
This is a standard integral of the form \( \int \frac{dx}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C \).
Here, \( a = \frac{9}{8} \) and \( u = x - \frac{9}{8} \).
So, \( I = \frac{1}{2} \sin^{-1}\left(\frac{x - \frac{9}{8}}{\frac{9}{8}}\right) + C \)
Simplify the argument of the inverse sine function:
\( \frac{x - \frac{9}{8}}{\frac{9}{8}} = \frac{\frac{8x-9}{8}}{\frac{9}{8}} = \frac{8x-9}{9} \)
Thus, the integral is \( I = \frac{1}{2} \sin^{-1}\left(\frac{8x-9}{9}\right) + C \).
This matches option (b).
In simple words: First, you factor out the negative sign and complete the square for the quadratic expression under the square root. This transforms the integral into a standard inverse sine form. Then, apply the specific formula for \( \int \frac{1}{\sqrt{a^2-x^2}} dx \) to get the final answer.

Exam Tip: Always be careful when completing the square, especially with negative coefficients for \( x^2 \). Factoring out the coefficient of \( x^2 \) first simplifies the process and helps avoid algebraic errors. Remember the exact forms for inverse trigonometric integrals.