OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral 3 Exercise 15 (E)

S Chand Class 12 ICSE Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(e)

Question 1. \( \int \frac{d x}{\sqrt{7-6 x-x^2}} \)
Answer: Let the integral be \( I = \int \frac{d x}{\sqrt{7-6 x-x^2}} \).
First, we rewrite the term inside the square root: \( 7-6x-x^2 = -\left(x^2+6x-7\right) \).
Next, complete the square for \( x^2+6x-7 \). We add and subtract \( \left(\frac{6}{2}\right)^2 = 9 \).
So, \( x^2+6x-7 = x^2+6x+9-9-7 = (x+3)^2-16 \).
Therefore, \( 7-6x-x^2 = -\left((x+3)^2-16\right) = 16-(x+3)^2 \).
The integral becomes \( I = \int \frac{d x}{\sqrt{16-(x+3)^2}} \).
Let \( x+3 = t \).
\( \implies \) Then \( dx = dt \).
So, \( I = \int \frac{d t}{\sqrt{4^2-t^2}} \).
This is a standard integral of the form \( \int \frac{du}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C \).
Here, \( a=4 \) and \( u=t \).
So, \( I = \sin^{-1}\left(\frac{t}{4}\right) + C \).
Substitute back \( t = x+3 \).
\( \implies I = \sin^{-1}\left(\frac{x+3}{4}\right) + C \).
In simple words: To solve this integral, we first rewrite the part under the square root by completing the square. This helps us change the expression into a simpler form. Then, we use a substitution to turn it into a known integral formula involving the sine inverse function. Finally, we put the original variable back into the answer.

🎯 Exam Tip: When dealing with integrals involving quadratic expressions under a square root, always try to complete the square to transform the quadratic into a standard form like \( a^2-x^2 \), \( x^2-a^2 \), or \( a^2+x^2 \), which can then be solved using standard formulas.

Question 2. \( \int \frac{d x}{\sqrt{10-8 x-2 x^2}} \)
Answer: Let the integral be \( I = \int \frac{d x}{\sqrt{10-8 x-2 x^2}} \).
First, take out -2 from the quadratic expression inside the square root: \( 10-8x-2x^2 = -2(x^2+4x-5) \).
So, \( I = \int \frac{d x}{\sqrt{-2(x^2+4x-5)}} = \frac{1}{\sqrt{2}}\int \frac{d x}{\sqrt{-(x^2+4x-5)}} \).
Now, complete the square for \( x^2+4x-5 \). We add and subtract \( \left(\frac{4}{2}\right)^2 = 4 \).
So, \( x^2+4x-5 = x^2+4x+4-4-5 = (x+2)^2-9 \).
Therefore, \( -(x^2+4x-5) = -\left((x+2)^2-9\right) = 9-(x+2)^2 \).
The integral becomes \( I = \frac{1}{\sqrt{2}}\int \frac{d x}{\sqrt{9-(x+2)^2}} \).
Let \( x+2 = t \).
\( \implies \) Then \( dx = dt \).
So, \( I = \frac{1}{\sqrt{2}}\int \frac{d t}{\sqrt{3^2-t^2}} \).
Using the standard formula \( \int \frac{du}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C \), where \( a=3 \) and \( u=t \).
\( \implies I = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{t}{3}\right) + C \).
Substitute back \( t = x+2 \).
\( \implies I = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{x+2}{3}\right) + C \).
In simple words: We simplify the expression under the square root by factoring out a constant and then completing the square. This transforms the integral into a known form. After a simple substitution, we can use the formula for sine inverse and then substitute back to get the final answer.

🎯 Exam Tip: Remember to factor out any coefficient of \( x^2 \) before completing the square to avoid errors in the transformation of the integral. The constant factor outside the square root must be handled carefully.

Question 3. \( \int \frac{d x}{\sqrt{4-2 x-2x^2}} \)
Answer: Let the integral be \( I = \int \frac{d x}{\sqrt{4-2 x-2x^2}} \).
First, factor out -2 from the quadratic expression: \( 4-2x-2x^2 = -2(x^2+x-2) \).
So, \( I = \int \frac{d x}{\sqrt{-2(x^2+x-2)}} = \frac{1}{\sqrt{2}}\int \frac{d x}{\sqrt{-(x^2+x-2)}} \).
Next, complete the square for \( x^2+x-2 \). Add and subtract \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
So, \( x^2+x-2 = x^2+x+\frac{1}{4}-\frac{1}{4}-2 = \left(x+\frac{1}{2}\right)^2-\frac{9}{4} \).
Then, \( -(x^2+x-2) = -\left(\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\right) = \frac{9}{4}-\left(x+\frac{1}{2}\right)^2 \).
The integral becomes \( I = \frac{1}{\sqrt{2}}\int \frac{d x}{\sqrt{\left(\frac{3}{2}\right)^2-\left(x+\frac{1}{2}\right)^2}} \).
Let \( x+\frac{1}{2} = t \).
\( \implies \) Then \( dx = dt \).
So, \( I = \frac{1}{\sqrt{2}}\int \frac{d t}{\sqrt{\left(\frac{3}{2}\right)^2-t^2}} \).
Using the standard formula \( \int \frac{du}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C \), where \( a=\frac{3}{2} \) and \( u=t \).
\( \implies I = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{t}{\frac{3}{2}}\right) + C = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{2t}{3}\right) + C \).
Substitute back \( t = x+\frac{1}{2} = \frac{2x+1}{2} \).
\( \implies I = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{2\left(\frac{2x+1}{2}\right)}{3}\right) + C = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{2x+1}{3}\right) + C \).
In simple words: This problem involves simplifying the expression under the square root by taking out a common factor and completing the square. Once it's in a standard form, we use a substitution and then apply the inverse sine integral formula. Remember to substitute back the original variable at the end.

🎯 Exam Tip: Pay close attention to the signs when factoring out coefficients and completing the square, especially when a negative sign is involved, as it determines which standard integral formula to use.

Question 4. \( \int \frac{d x}{\sqrt{16-2 x-2 x^2}} \)
Answer: Let the integral be \( I = \int \frac{d x}{\sqrt{16-2 x-2 x^2}} \).
First, factor out -2 from the quadratic expression: \( 16-2x-2x^2 = -2(x^2+x-8) \).
So, \( I = \int \frac{d x}{\sqrt{-2(x^2+x-8)}} = \frac{1}{\sqrt{2}}\int \frac{d x}{\sqrt{-(x^2+x-8)}} \).
Next, complete the square for \( x^2+x-8 \). Add and subtract \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
So, \( x^2+x-8 = x^2+x+\frac{1}{4}-\frac{1}{4}-8 = \left(x+\frac{1}{2}\right)^2-\frac{33}{4} \).
Then, \( -(x^2+x-8) = -\left(\left(x+\frac{1}{2}\right)^2-\frac{33}{4}\right) = \frac{33}{4}-\left(x+\frac{1}{2}\right)^2 \).
The integral becomes \( I = \frac{1}{\sqrt{2}}\int \frac{d x}{\sqrt{\left(\frac{\sqrt{33}}{2}\right)^2-\left(x+\frac{1}{2}\right)^2}} \).
Let \( x+\frac{1}{2} = t \).
\( \implies \) Then \( dx = dt \).
So, \( I = \frac{1}{\sqrt{2}}\int \frac{d t}{\sqrt{\left(\frac{\sqrt{33}}{2}\right)^2-t^2}} \).
Using the standard formula \( \int \frac{du}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C \), where \( a=\frac{\sqrt{33}}{2} \) and \( u=t \).
\( \implies I = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{t}{\frac{\sqrt{33}}{2}}\right) + C = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{2t}{\sqrt{33}}\right) + C \).
Substitute back \( t = x+\frac{1}{2} = \frac{2x+1}{2} \).
\( \implies I = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{2\left(\frac{2x+1}{2}\right)}{\sqrt{33}}\right) + C = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{2x+1}{\sqrt{33}}\right) + C \).
In simple words: This solution involves algebraic manipulation to simplify the quadratic expression under the square root. We factor out the coefficient of \( x^2 \) and then complete the square to get it into a form that matches a known integral identity. A simple substitution and then applying the inverse sine formula helps find the final integral.

🎯 Exam Tip: When the coefficient of \( x^2 \) is not 1, always factor it out first. Remember to handle the square root of this factored number correctly outside the integral.

Question 5. \( \int \frac{e^x}{\sqrt{5-4 e^x-e^{2 x}}} d x \)
Answer: Let the integral be \( I = \int \frac{e^x}{\sqrt{5-4 e^x-e^{2 x}}} d x \).
First, let's use a substitution. Let \( t = e^x \).
\( \implies \) Then \( dt = e^x dx \). Also, \( e^{2x} = (e^x)^2 = t^2 \).
The integral becomes \( I = \int \frac{d t}{\sqrt{5-4 t-t^2}} \).
Next, rewrite the term inside the square root by factoring out a negative sign: \( 5-4t-t^2 = -\left(t^2+4t-5\right) \).
Now, complete the square for \( t^2+4t-5 \). We add and subtract \( \left(\frac{4}{2}\right)^2 = 4 \).
So, \( t^2+4t-5 = t^2+4t+4-4-5 = (t+2)^2-9 \).
Therefore, \( -\left(t^2+4t-5\right) = -\left((t+2)^2-9\right) = 9-(t+2)^2 \).
The integral becomes \( I = \int \frac{d t}{\sqrt{9-(t+2)^2}} \).
Let \( u = t+2 \).
\( \implies \) Then \( du = dt \).
So, \( I = \int \frac{d u}{\sqrt{3^2-u^2}} \).
Using the standard formula \( \int \frac{du}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C \), where \( a=3 \) and \( u=u \).
\( \implies I = \sin^{-1}\left(\frac{u}{3}\right) + C \).
Substitute back \( u = t+2 \).
\( \implies I = \sin^{-1}\left(\frac{t+2}{3}\right) + C \).
Finally, substitute back \( t = e^x \).
\( \implies I = \sin^{-1}\left(\frac{e^x+2}{3}\right) + C \).
In simple words: First, we use a substitution to change the \( e^x \) terms into a simpler variable. Then, we manipulate the expression under the square root by completing the square, making it match a standard integral form. After another substitution and applying the inverse sine formula, we replace the variables to get the answer in terms of \( x \).

🎯 Exam Tip: For integrals involving exponential functions like \( e^x \) and \( e^{2x} \), a common strategy is to substitute \( t=e^x \) to simplify the expression, often leading to a standard integral form.

Question 6. \( \int \frac{1}{\sqrt{(x-1)(x-2)}} d x \)
Answer: Let the integral be \( I = \int \frac{d x}{\sqrt{(x-1)(x-2)}} \).
First, expand the expression in the denominator: \( (x-1)(x-2) = x^2-2x-x+2 = x^2-3x+2 \).
So, \( I = \int \frac{d x}{\sqrt{x^2-3x+2}} \).
Now, complete the square for \( x^2-3x+2 \). We add and subtract \( \left(\frac{-3}{2}\right)^2 = \frac{9}{4} \).
So, \( x^2-3x+2 = x^2-3x+\frac{9}{4}-\frac{9}{4}+2 = \left(x-\frac{3}{2}\right)^2-\frac{1}{4} \).
The integral becomes \( I = \int \frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}} \).
This is a standard integral of the form \( \int \frac{du}{\sqrt{u^2-a^2}} = \log\left|u+\sqrt{u^2-a^2}\right| + C \).
Here, \( u=x-\frac{3}{2} \) and \( a=\frac{1}{2} \).
\( \implies I = \log\left|x-\frac{3}{2}+\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right| + C \).
Substitute back the original quadratic expression: \( \left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2 = x^2-3x+2 \).
\( \implies I = \log\left|x-\frac{3}{2}+\sqrt{x^2-3x+2}\right| + C \).
In simple words: We first multiply out the terms under the square root, then complete the square to get the expression into a specific form. This form matches a known integration formula involving logarithms. By applying that formula, we find the answer.

🎯 Exam Tip: When the denominator has a product of linear factors under a square root, expand it first and then complete the square. This will likely lead to an integral of the form \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} \).

Question 7. \( \int \frac{1}{\sqrt{5 x^2-2 x}} d x \)
Answer: Let the integral be \( I = \int \frac{d x}{\sqrt{5 x^2-2 x}} \).
First, factor out the coefficient of \( x^2 \) from under the square root: \( 5x^2-2x = 5\left(x^2-\frac{2}{5}x\right) \).
So, \( I = \int \frac{d x}{\sqrt{5\left(x^2-\frac{2}{5}x\right)}} = \frac{1}{\sqrt{5}}\int \frac{d x}{\sqrt{x^2-\frac{2}{5}x}} \).
Now, complete the square for \( x^2-\frac{2}{5}x \). We add and subtract \( \left(\frac{1}{2} \cdot \frac{-2}{5}\right)^2 = \left(-\frac{1}{5}\right)^2 = \frac{1}{25} \).
So, \( x^2-\frac{2}{5}x = x^2-\frac{2}{5}x+\frac{1}{25}-\frac{1}{25} = \left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2 \).
The integral becomes \( I = \frac{1}{\sqrt{5}}\int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}} \).
This is a standard integral of the form \( \int \frac{du}{\sqrt{u^2-a^2}} = \log\left|u+\sqrt{u^2-a^2}\right| + C \).
Here, \( u=x-\frac{1}{5} \) and \( a=\frac{1}{5} \).
\( \implies I = \frac{1}{\sqrt{5}}\log\left|x-\frac{1}{5}+\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}\right| + C \).
Substitute back the original quadratic expression: \( \left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2 = x^2-\frac{2}{5}x \).
\( \implies I = \frac{1}{\sqrt{5}}\log\left|x-\frac{1}{5}+\sqrt{x^2-\frac{2}{5}x}\right| + C \).
To simplify the term inside the log, we can multiply \( \sqrt{x^2-\frac{2}{5}x} \) by \( \sqrt{5} \) to make the expression look cleaner, but the form \( \sqrt{x^2-\frac{2}{5}x} \) is also correct.
Alternatively, \( I = \frac{1}{\sqrt{5}}\log\left|\frac{5x-1}{5}+\frac{\sqrt{25x^2-10x}}{5}\right| + C \).
\( \implies I = \frac{1}{\sqrt{5}}\log\left|\frac{5x-1+\sqrt{25x^2-10x}}{5}\right| + C \).
The constant \( -\frac{1}{\sqrt{5}}\log 5 \) can be absorbed into C.
So, \( I = \frac{1}{\sqrt{5}}\log\left|5x-1+\sqrt{25x^2-10x}\right| + C \).
In simple words: First, we take out the number in front of \( x^2 \) from under the square root. Then, we complete the square for the remaining part to get it into a standard form. This lets us use a known integration rule that involves a logarithm to find the answer.

🎯 Exam Tip: When \( x^2 \) has a coefficient, factor it out carefully from the entire expression under the square root. Remember to apply the square root to this factored coefficient and place it outside the integral.

Question 8. \( \int \frac{2 x+1}{\sqrt{x^2+2 x-1}} d x \)
Answer: Let the integral be \( I = \int \frac{2 x+1}{\sqrt{x^2+2 x-1}} d x \).
We can split the numerator to handle this integral. We want a term like \( 2x+2 \) (the derivative of \( x^2+2x-1 \)).
So, \( 2x+1 = (2x+2)-1 \).
\( \implies I = \int \frac{(2 x+2)-1}{\sqrt{x^2+2 x-1}} d x = \int \frac{2 x+2}{\sqrt{x^2+2 x-1}} d x - \int \frac{1}{\sqrt{x^2+2 x-1}} d x \).
Let's call these \( I_1 \) and \( I_2 \). So, \( I = I_1 - I_2 \).
For \( I_1 = \int \frac{2 x+2}{\sqrt{x^2+2 x-1}} d x \):
Let \( t = x^2+2x-1 \).
\( \implies \) Then \( dt = (2x+2)dx \).
So, \( I_1 = \int \frac{d t}{\sqrt{t}} = \int t^{-\frac{1}{2}} d t \).
\( \implies I_1 = \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + C_1 = \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + C_1 = 2\sqrt{t} + C_1 = 2\sqrt{x^2+2x-1} + C_1 \).
For \( I_2 = \int \frac{1}{\sqrt{x^2+2 x-1}} d x \):
Complete the square for \( x^2+2x-1 \). Add and subtract \( \left(\frac{2}{2}\right)^2 = 1 \).
So, \( x^2+2x-1 = x^2+2x+1-1-1 = (x+1)^2-2 \).
\( \implies I_2 = \int \frac{d x}{\sqrt{(x+1)^2-(\sqrt{2})^2}} \).
This is a standard integral of the form \( \int \frac{du}{\sqrt{u^2-a^2}} = \log\left|u+\sqrt{u^2-a^2}\right| + C \).
Here, \( u=x+1 \) and \( a=\sqrt{2} \).
\( \implies I_2 = \log\left|x+1+\sqrt{(x+1)^2-(\sqrt{2})^2}\right| + C_2 \).
Substitute back \( (x+1)^2-(\sqrt{2})^2 = x^2+2x-1 \).
\( \implies I_2 = \log\left|x+1+\sqrt{x^2+2x-1}\right| + C_2 \).
Combining \( I_1 \) and \( I_2 \):
\( I = I_1 - I_2 = 2\sqrt{x^2+2x-1} - \log\left|x+1+\sqrt{x^2+2x-1}\right| + C \).
In simple words: We split the integral into two parts. For the first part, we notice that the numerator is almost the derivative of the term under the square root, so a simple substitution works. For the second part, we complete the square in the denominator to use a standard logarithmic integral formula. Finally, we combine these two results.

🎯 Exam Tip: When the numerator is a linear function and the denominator is a square root of a quadratic, try to express the numerator as \( A \cdot (\text{derivative of quadratic}) + B \) to split the integral into two solvable parts.

Question 9. \( \int \frac{x d x}{\sqrt{8+x-x^2}} \)
Answer: Let the integral be \( I = \int \frac{x d x}{\sqrt{8+x-x^2}} \).
We need to express the numerator \( x \) in terms of the derivative of \( 8+x-x^2 \).
The derivative of \( 8+x-x^2 \) is \( 1-2x \).
Let \( x = A(1-2x) + B \).
\( x = A - 2Ax + B \).
Comparing coefficients of \( x \): \( 1 = -2A \implies A = -\frac{1}{2} \).
Comparing constant terms: \( 0 = A+B \implies B = -A = \frac{1}{2} \).
So, \( x = -\frac{1}{2}(1-2x) + \frac{1}{2} \).
\( \implies I = \int \frac{-\frac{1}{2}(1-2x) + \frac{1}{2}}{\sqrt{8+x-x^2}} d x = -\frac{1}{2}\int \frac{1-2x}{\sqrt{8+x-x^2}} d x + \frac{1}{2}\int \frac{1}{\sqrt{8+x-x^2}} d x \).
Let's call these \( I_1 \) and \( I_2 \). So, \( I = -\frac{1}{2}I_1 + \frac{1}{2}I_2 \).
For \( I_1 = \int \frac{1-2x}{\sqrt{8+x-x^2}} d x \):
Let \( t = 8+x-x^2 \).
\( \implies \) Then \( dt = (1-2x)dx \).
So, \( I_1 = \int \frac{d t}{\sqrt{t}} = \int t^{-\frac{1}{2}} d t = 2\sqrt{t} + C_1 = 2\sqrt{8+x-x^2} + C_1 \).
For \( I_2 = \int \frac{1}{\sqrt{8+x-x^2}} d x \):
Complete the square for \( 8+x-x^2 = -\left(x^2-x-8\right) \). Add and subtract \( \left(\frac{-1}{2}\right)^2 = \frac{1}{4} \).
So, \( x^2-x-8 = x^2-x+\frac{1}{4}-\frac{1}{4}-8 = \left(x-\frac{1}{2}\right)^2-\frac{33}{4} \).
Therefore, \( 8+x-x^2 = -\left(\left(x-\frac{1}{2}\right)^2-\frac{33}{4}\right) = \frac{33}{4}-\left(x-\frac{1}{2}\right)^2 \).
\( \implies I_2 = \int \frac{d x}{\sqrt{\left(\frac{\sqrt{33}}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}} \).
Using the standard formula \( \int \frac{du}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C \).
Here, \( u=x-\frac{1}{2} \) and \( a=\frac{\sqrt{33}}{2} \).
\( \implies I_2 = \sin^{-1}\left(\frac{x-\frac{1}{2}}{\frac{\sqrt{33}}{2}}\right) + C_2 = \sin^{-1}\left(\frac{2x-1}{\sqrt{33}}\right) + C_2 \).
Combining the results:
\( I = -\frac{1}{2}(2\sqrt{8+x-x^2}) + \frac{1}{2}\left(\sin^{-1}\left(\frac{2x-1}{\sqrt{33}}\right)\right) + C \).
\( \implies I = -\sqrt{8+x-x^2} + \frac{1}{2}\sin^{-1}\left(\frac{2x-1}{\sqrt{33}}\right) + C \).
In simple words: To solve this, we adjust the numerator so it includes the derivative of the expression under the square root. This splits the integral into two simpler parts. The first part is solved by a direct substitution. The second part requires completing the square and using the inverse sine formula. Finally, we combine both results to get the full answer.

🎯 Exam Tip: For integrals where the numerator is a linear function and the denominator is the square root of a quadratic, always express the numerator as a linear combination of the derivative of the quadratic and a constant. This technique is key to breaking down the integral.

Question 10. \( \int \frac{6 x+7}{\sqrt{(x-5)(x-4)}} d x \)
Answer: Let the integral be \( I = \int \frac{6 x+7}{\sqrt{(x-5)(x-4)}} d x \).
First, expand the term in the denominator: \( (x-5)(x-4) = x^2-4x-5x+20 = x^2-9x+20 \).
So, \( I = \int \frac{6 x+7}{\sqrt{x^2-9x+20}} d x \).
We need to express the numerator \( 6x+7 \) in terms of the derivative of \( x^2-9x+20 \).
The derivative of \( x^2-9x+20 \) is \( 2x-9 \).
Let \( 6x+7 = A(2x-9) + B \).
\( 6x+7 = 2Ax - 9A + B \).
Comparing coefficients of \( x \): \( 6 = 2A \implies A = 3 \).
Comparing constant terms: \( 7 = -9A+B \implies 7 = -9(3)+B \implies 7 = -27+B \implies B = 34 \).
So, \( 6x+7 = 3(2x-9) + 34 \).
\( \implies I = \int \frac{3(2x-9) + 34}{\sqrt{x^2-9x+20}} d x = 3\int \frac{2x-9}{\sqrt{x^2-9x+20}} d x + 34\int \frac{1}{\sqrt{x^2-9x+20}} d x \).
Let's call these \( I_1 \) and \( I_2 \). So, \( I = 3I_1 + 34I_2 \).
For \( I_1 = \int \frac{2x-9}{\sqrt{x^2-9x+20}} d x \):
Let \( t = x^2-9x+20 \).
\( \implies \) Then \( dt = (2x-9)dx \).
So, \( I_1 = \int \frac{d t}{\sqrt{t}} = \int t^{-\frac{1}{2}} d t = 2\sqrt{t} + C_1 = 2\sqrt{x^2-9x+20} + C_1 \).
For \( I_2 = \int \frac{1}{\sqrt{x^2-9x+20}} d x \):
Complete the square for \( x^2-9x+20 \). Add and subtract \( \left(\frac{-9}{2}\right)^2 = \frac{81}{4} \).
So, \( x^2-9x+20 = x^2-9x+\frac{81}{4}-\frac{81}{4}+20 = \left(x-\frac{9}{2}\right)^2-\frac{1}{4} \).
\( \implies I_2 = \int \frac{d x}{\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2}} \).
Using the standard formula \( \int \frac{du}{\sqrt{u^2-a^2}} = \log\left|u+\sqrt{u^2-a^2}\right| + C \).
Here, \( u=x-\frac{9}{2} \) and \( a=\frac{1}{2} \).
\( \implies I_2 = \log\left|x-\frac{9}{2}+\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right| + C_2 \).
Substitute back \( \left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2 = x^2-9x+20 \).
\( \implies I_2 = \log\left|x-\frac{9}{2}+\sqrt{x^2-9x+20}\right| + C_2 \).
Combining the results:
\( I = 3(2\sqrt{x^2-9x+20}) + 34\left(\log\left|x-\frac{9}{2}+\sqrt{x^2-9x+20}\right|\right) + C \).
\( \implies I = 6\sqrt{x^2-9x+20} + 34\log\left|x-\frac{9}{2}+\sqrt{x^2-9x+20}\right| + C \).
In simple words: First, we expand the denominator and then rewrite the numerator so it includes the derivative of the expression under the square root. This splits the integral into two parts. The first part is solved by direct substitution, and the second part requires completing the square and using a logarithmic integral formula. Finally, we combine both answers.

🎯 Exam Tip: When faced with an integral of a linear function divided by the square root of a quadratic, always start by expressing the numerator as \( A \cdot (\text{derivative of quadratic}) + B \). This method systematically breaks the integral into simpler forms.

Question 11. \( \int \frac{4 x+1}{\sqrt{2 x^2+x-3}} d x \)
Answer: Let the integral be \( I = \int \frac{4 x+1}{\sqrt{2 x^2+x-3}} d x \).
Let's observe the relationship between the numerator and the derivative of the term inside the square root.
Let \( t = 2x^2+x-3 \).
\( \implies \) Then \( dt = (4x+1)dx \).
The numerator \( 4x+1 \) is exactly the derivative of \( 2x^2+x-3 \).
So, the integral becomes \( I = \int \frac{d t}{\sqrt{t}} \).
This can be written as \( I = \int t^{-\frac{1}{2}} d t \).
Using the power rule for integration, \( \int u^n du = \frac{u^{n+1}}{n+1} + C \).
\( \implies I = \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + C = \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + C = 2\sqrt{t} + C \).
Substitute back \( t = 2x^2+x-3 \).
\( \implies I = 2\sqrt{2x^2+x-3} + C \).
In simple words: We notice that the top part of the fraction is exactly the derivative of the expression under the square root. This means we can use a simple substitution. After making the substitution, the integral becomes very easy to solve using the power rule, and then we just put the original expression back.

🎯 Exam Tip: Always check if the numerator is directly the derivative of the expression under the square root in the denominator. If it is, a simple substitution \( u = \text{denominator expression} \) will greatly simplify the integral.

Question 12. \( \int \frac{x}{\sqrt{x^2+x+1}} d x \)
Answer: Let the integral be \( I = \int \frac{x}{\sqrt{x^2+x+1}} d x \).
The derivative of \( x^2+x+1 \) is \( 2x+1 \). We need to transform the numerator \( x \) to involve \( 2x+1 \).
We can write \( x = \frac{1}{2}(2x+1) - \frac{1}{2} \).
\( \implies I = \int \frac{\frac{1}{2}(2x+1) - \frac{1}{2}}{\sqrt{x^2+x+1}} d x = \frac{1}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}} d x - \frac{1}{2}\int \frac{1}{\sqrt{x^2+x+1}} d x \).
Let's call these \( I_1 \) and \( I_2 \). So, \( I = \frac{1}{2}I_1 - \frac{1}{2}I_2 \).
For \( I_1 = \int \frac{2x+1}{\sqrt{x^2+x+1}} d x \):
Let \( t = x^2+x+1 \).
\( \implies \) Then \( dt = (2x+1)dx \).
So, \( I_1 = \int \frac{d t}{\sqrt{t}} = \int t^{-\frac{1}{2}} d t = 2\sqrt{t} + C_1 = 2\sqrt{x^2+x+1} + C_1 \).
For \( I_2 = \int \frac{1}{\sqrt{x^2+x+1}} d x \):
Complete the square for \( x^2+x+1 \). Add and subtract \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
So, \( x^2+x+1 = x^2+x+\frac{1}{4}-\frac{1}{4}+1 = \left(x+\frac{1}{2}\right)^2+\frac{3}{4} \).
\( \implies I_2 = \int \frac{d x}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}} \).
This is a standard integral of the form \( \int \frac{du}{\sqrt{u^2+a^2}} = \log\left|u+\sqrt{u^2+a^2}\right| + C \).
Here, \( u=x+\frac{1}{2} \) and \( a=\frac{\sqrt{3}}{2} \).
\( \implies I_2 = \log\left|x+\frac{1}{2}+\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right| + C_2 \).
Substitute back \( \left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2 = x^2+x+1 \).
\( \implies I_2 = \log\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right| + C_2 \).
Combining the results:
\( I = \frac{1}{2}(2\sqrt{x^2+x+1}) - \frac{1}{2}\left(\log\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|\right) + C \).
\( \implies I = \sqrt{x^2+x+1} - \frac{1}{2}\log\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right| + C \).
In simple words: We rewrite the numerator to separate the integral into two parts: one where the numerator is the derivative of the expression under the square root, and another simple integral. The first part is solved by a quick substitution. For the second part, we complete the square and use a standard logarithmic formula. Finally, we combine these two answers.

🎯 Exam Tip: When the numerator is a linear term and the denominator contains a quadratic under a square root, always try to express the numerator in terms of the derivative of the quadratic, which helps split the integral into manageable parts.

Question 13. \( \int \frac{x+2}{\sqrt{x^2-1}} d x \)
Answer: Let the integral be \( I = \int \frac{x+2}{\sqrt{x^2-1}} d x \).
We can split the numerator to solve this integral.
\( \implies I = \int \frac{x}{\sqrt{x^2-1}} d x + \int \frac{2}{\sqrt{x^2-1}} d x \).
Let's call these \( I_1 \) and \( I_2 \). So, \( I = I_1 + I_2 \).
For \( I_1 = \int \frac{x}{\sqrt{x^2-1}} d x \):
Let \( t = x^2-1 \).
\( \implies \) Then \( dt = 2x dx \implies x dx = \frac{1}{2} dt \).
So, \( I_1 = \int \frac{\frac{1}{2} d t}{\sqrt{t}} = \frac{1}{2}\int t^{-\frac{1}{2}} d t \).
\( \implies I_1 = \frac{1}{2}\left(\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right) + C_1 = \frac{1}{2}\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right) + C_1 = \sqrt{t} + C_1 = \sqrt{x^2-1} + C_1 \).
For \( I_2 = \int \frac{2}{\sqrt{x^2-1}} d x \):
\( I_2 = 2\int \frac{d x}{\sqrt{x^2-1^2}} \).
This is a standard integral of the form \( \int \frac{du}{\sqrt{u^2-a^2}} = \log\left|u+\sqrt{u^2-a^2}\right| + C \).
Here, \( u=x \) and \( a=1 \).
\( \implies I_2 = 2\log\left|x+\sqrt{x^2-1^2}\right| + C_2 = 2\log\left|x+\sqrt{x^2-1}\right| + C_2 \).
Combining the results:
\( I = \sqrt{x^2-1} + 2\log\left|x+\sqrt{x^2-1}\right| + C \).
In simple words: We split the integral into two parts by separating the numerator. For the first part, a simple substitution helps because the numerator is related to the derivative of the term under the square root. For the second part, we use a known integration formula involving logarithms. Finally, we add both parts to get the complete answer.

🎯 Exam Tip: If the numerator is a sum of terms and the denominator is a single expression, split the integral into multiple parts. This often simplifies each part into a standard integrable form.

Question 14. \( \int \sqrt{\frac{5-x}{x-2}} d x \)
Answer: The solution steps for Question 14 are not available within the provided page range (pages 1-14). The full mathematical working for this integral appears on subsequent pages which are outside the scope of this processing task.
In simple words: The complete solution for this question cannot be provided because the necessary mathematical steps are on pages not included in the allowed content.

🎯 Exam Tip: For integrals of the form \( \int \sqrt{\frac{a-x}{b-x}} dx \) or \( \int \sqrt{\frac{a+x}{b+x}} dx \), rationalize the numerator by multiplying both numerator and denominator by \( \sqrt{a-x} \) (or \( \sqrt{a+x} \)). This converts the integral into a form that can be solved by completing the square and using standard formulas.

Question 14. \( \int \sqrt{\frac{5-x}{x-2}} d x \)
Answer: Let the integral be \( I \). First, we rewrite the expression under the square root.
\( I = \int \sqrt{\frac{5-x}{x-2}} d x \)
To solve this, we multiply the numerator and denominator by \( \sqrt{5-x} \).
\( I = \int \frac{5-x}{\sqrt{(x-2)(5-x)}} d x \)
Next, expand the terms in the denominator.
\( I = \int \frac{5-x}{\sqrt{-x^2+7x-10}} d x \)
We want to make the numerator relate to the derivative of the expression inside the square root in the denominator. Let \( f(x) = -x^2+7x-10 \), so \( f'(x) = -2x+7 \). We express \( 5-x \) in terms of \( f'(x) \).
\( 5-x = \frac{1}{2}(-2x+7) + \frac{3}{2} \)
Now, substitute this into the integral and split it into two parts.
\( I = \frac{1}{2} \int \frac{-2x+7}{\sqrt{-x^2+7x-10}} d x + \frac{3}{2} \int \frac{1}{\sqrt{-x^2+7x-10}} d x \)
For the first integral, use the substitution \( t = -x^2+7x-10 \), so \( dt = (-2x+7)dx \).
\( \frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} (2\sqrt{t}) = \sqrt{t} = \sqrt{-x^2+7x-10} \)
For the second integral, complete the square for the quadratic expression in the denominator.
\( -x^2+7x-10 = - (x^2-7x+10) = - (x^2-7x + (\frac{7}{2})^2 - (\frac{7}{2})^2 + 10) \)
\( = - ( (x-\frac{7}{2})^2 - \frac{49}{4} + \frac{40}{4} ) = - ( (x-\frac{7}{2})^2 - \frac{9}{4} ) = \frac{9}{4} - (x-\frac{7}{2})^2 \)
So, the second integral becomes:
\( \frac{3}{2} \int \frac{1}{\sqrt{(\frac{3}{2})^2 - (x-\frac{7}{2})^2}} d x \)
This is of the form \( \int \frac{1}{\sqrt{a^2-y^2}} dy = \sin^{-1}(\frac{y}{a}) \). Here \( a = \frac{3}{2} \) and \( y = x-\frac{7}{2} \).
\( \frac{3}{2} \sin^{-1} \left( \frac{x-\frac{7}{2}}{\frac{3}{2}} \right) + C = \frac{3}{2} \sin^{-1} \left( \frac{2x-7}{3} \right) + C \)
Combining both parts, the final solution is:
\( I = \sqrt{-x^2+7x-10} + \frac{3}{2} \sin^{-1} \left( \frac{2x-7}{3} \right) + C \)
In simple words: We first changed the fraction inside the square root to make it easier to work with. Then, we split the problem into two smaller parts. For the first part, we used a simple substitution. For the second part, we used a method called completing the square to simplify the expression and solve it using a standard integral formula for sine inverse.

🎯 Exam Tip: When you have an integral of the form \( \int \frac{Px+Q}{\sqrt{Ax^2+Bx+C}} dx \), always write the numerator \( Px+Q \) as \( A'(2Ax+B) + B' \) to split the integral into two standard forms.

Question 15. \( \int \frac{5 x+3}{\sqrt{x^2+4x+10}} d x \)
Answer: Let the integral be \( I \). This is an integral of the form \( \int \frac{Px+Q}{\sqrt{Ax^2+Bx+C}} dx \).
We write the numerator \( 5x+3 \) in terms of the derivative of \( x^2+4x+10 \), which is \( 2x+4 \).
Let \( 5x+3 = A(2x+4) + B \).
Comparing coefficients of \( x \): \( 5 = 2A \implies A = \frac{5}{2} \).
Comparing constant terms: \( 3 = 4A + B = 4(\frac{5}{2}) + B = 10 + B \implies B = 3-10 = -7 \).
So, \( 5x+3 = \frac{5}{2}(2x+4) - 7 \).
Substitute this back into the integral and split it into two parts.
\( I = \int \frac{\frac{5}{2}(2x+4) - 7}{\sqrt{x^2+4x+10}} d x \)
\( I = \frac{5}{2} \int \frac{2x+4}{\sqrt{x^2+4x+10}} d x - 7 \int \frac{1}{\sqrt{x^2+4x+10}} d x \)
For the first integral, let \( t = x^2+4x+10 \). Then \( dt = (2x+4)dx \).
\( \frac{5}{2} \int \frac{dt}{\sqrt{t}} = \frac{5}{2} (2\sqrt{t}) = 5\sqrt{t} = 5\sqrt{x^2+4x+10} \)
For the second integral, we complete the square for the expression \( x^2+4x+10 \).
\( x^2+4x+10 = x^2+4x+4+6 = (x+2)^2 + 6 = (x+2)^2 + (\sqrt{6})^2 \)
So, the second integral becomes:
\( -7 \int \frac{1}{\sqrt{(x+2)^2 + (\sqrt{6})^2}} d x \)
This is of the form \( \int \frac{1}{\sqrt{y^2+a^2}} dy = \log |y+\sqrt{y^2+a^2}| \). Here \( y=x+2 \) and \( a=\sqrt{6} \).
\( -7 \log |(x+2) + \sqrt{(x+2)^2+6}| + C = -7 \log |x+2 + \sqrt{x^2+4x+10}| + C \)
Combining the results from both parts, the final solution is:
\( I = 5\sqrt{x^2+4x+10} - 7 \log |x+2 + \sqrt{x^2+4x+10}| + C \)
In simple words: We changed the top part of the fraction to match the derivative of the bottom part. This let us break the problem into two standard integrals. We solved the first part using a simple substitution, and the second part by completing the square and using a logarithm formula.

🎯 Exam Tip: Remember to express the numerator as a linear combination of the derivative of the quadratic expression in the denominator and a constant. This is a common strategy for such integral types.

Question 16. \( \int \frac{x+2}{\sqrt{x^2+5x+6}} d x \)
Answer: Let the integral be \( I \). This is also of the form \( \int \frac{Px+Q}{\sqrt{Ax^2+Bx+C}} dx \).
We write the numerator \( x+2 \) in terms of the derivative of \( x^2+5x+6 \), which is \( 2x+5 \).
Let \( x+2 = A(2x+5) + B \).
Comparing coefficients of \( x \): \( 1 = 2A \implies A = \frac{1}{2} \).
Comparing constant terms: \( 2 = 5A + B = 5(\frac{1}{2}) + B = \frac{5}{2} + B \implies B = 2-\frac{5}{2} = -\frac{1}{2} \).
So, \( x+2 = \frac{1}{2}(2x+5) - \frac{1}{2} \).
Substitute this back into the integral and split it into two parts.
\( I = \int \frac{\frac{1}{2}(2x+5) - \frac{1}{2}}{\sqrt{x^2+5x+6}} d x \)
\( I = \frac{1}{2} \int \frac{2x+5}{\sqrt{x^2+5x+6}} d x - \frac{1}{2} \int \frac{1}{\sqrt{x^2+5x+6}} d x \)
For the first integral, let \( t = x^2+5x+6 \). Then \( dt = (2x+5)dx \).
\( \frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} (2\sqrt{t}) = \sqrt{t} = \sqrt{x^2+5x+6} \)
For the second integral, we complete the square for the expression \( x^2+5x+6 \).
\( x^2+5x+6 = x^2+5x+(\frac{5}{2})^2 - (\frac{5}{2})^2 + 6 \)
\( = (x+\frac{5}{2})^2 - \frac{25}{4} + \frac{24}{4} = (x+\frac{5}{2})^2 - \frac{1}{4} = (x+\frac{5}{2})^2 - (\frac{1}{2})^2 \)
So, the second integral becomes:
\( -\frac{1}{2} \int \frac{1}{\sqrt{(x+\frac{5}{2})^2 - (\frac{1}{2})^2}} d x \)
This is of the form \( \int \frac{1}{\sqrt{y^2-a^2}} dy = \log |y+\sqrt{y^2-a^2}| \). Here \( y=x+\frac{5}{2} \) and \( a=\frac{1}{2} \).
\( -\frac{1}{2} \log |(x+\frac{5}{2}) + \sqrt{(x+\frac{5}{2})^2 - (\frac{1}{2})^2}| + C \)
\( = -\frac{1}{2} \log |x+\frac{5}{2} + \sqrt{x^2+5x+6}| + C \)
Combining the results from both parts, the final solution is:
\( I = \sqrt{x^2+5x+6} - \frac{1}{2} \log |x+\frac{5}{2} + \sqrt{x^2+5x+6}| + C \)
In simple words: We adjusted the top part of the fraction to match the derivative of the expression under the square root in the bottom part. This allowed us to split the integral into two simpler integrals. One part was solved by a direct substitution, and the other by completing the square and using a standard logarithm integral formula.

🎯 Exam Tip: Pay close attention to the signs when completing the square, especially when the \( x^2 \) term is negative, as this affects the choice of the standard integral formula.