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

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

Question 1. Write the integral for each of the following.
(i) \( \int \frac{dx}{x^2+2x+10} \)
(ii) \( \int \frac{dx}{(x+1)(x+2)} \)
(iii) \( \int \frac{d x}{9 x^2+6 x+1} \)
Answer:
(i) Let the integral be \( I = \int \frac{d x}{x^2+2x+10} \)
First, we complete the square in the denominator: \( x^2+2x+10 = x^2+2x+1+9 = (x+1)^2+3^2 \).
So, \( I = \int \frac{d x}{(x+1)^2+3^2} \)
Now, let's substitute \( x+1 = t \), which means \( dx = dt \).
\( I = \int \frac{d t}{t^2+3^2} \)
Using the formula \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C \), we get:
\( I = \frac{1}{3}\tan^{-1}\left(\frac{t}{3}\right)+C \)
Substitute back \( t = x+1 \):
\( I = \frac{1}{3}\tan^{-1}\left(\frac{x+1}{3}\right)+C \)

(ii) Let the integral be \( I = \int \frac{d x}{(x+1)(x+2)} \)
We can multiply the terms in the denominator: \( (x+1)(x+2) = x^2+3x+2 \).
So, \( I = \int \frac{d x}{x^2+3 x+2} \)
Now, complete the square in the denominator: \( x^2+3x+2 = x^2+3x+\frac{9}{4}-\frac{9}{4}+2 = \left(x+\frac{3}{2}\right)^2 - \frac{1}{4} = \left(x+\frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \).
So, \( I = \int \frac{d x}{\left(x+\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
Using the formula \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C \), with \( x = x+\frac{3}{2} \) and \( a = \frac{1}{2} \):
\( I = \frac{1}{2 \times \frac{1}{2}} \log \left|\frac{x+\frac{3}{2}-\frac{1}{2}}{x+\frac{3}{2}+\frac{1}{2}}\right|+C \)
Simplify the expression inside the logarithm:
\( I = \log \left|\frac{x+1}{x+2}\right|+C \)

(iii) Let the integral be \( I = \int \frac{d x}{9 x^2+6 x+1} \)
The denominator is a perfect square: \( 9x^2+6x+1 = (3x+1)^2 \).
So, \( I = \int \frac{d x}{(3x+1)^2} = \int (3x+1)^{-2} d x \)
Now, we can integrate this using the power rule \( \int u^n du = \frac{u^{n+1}}{n+1} \). Remember to divide by the derivative of the inner function (3 in this case).
\( I = \frac{(3x+1)^{-2+1}}{(-2+1) \times 3} + C \)
\( I = \frac{(3x+1)^{-1}}{-1 \times 3} + C \)
\( I = -\frac{1}{3(3x+1)} + C \)
In simple words: For each problem, we changed the bottom part of the fraction into a simpler form, like a squared term plus a number, or a perfect square. Then, we used special rules for integration to find the answer. It's like finding the reverse of differentiation.

🎯 Exam Tip: Always look for ways to complete the square or factorize the denominator to match standard integration formulas. Don't forget the constant of integration \(+C\).

Question 2. Find the integral for each expression.
(i) \( \int \frac{d x}{9 x^2+6 x+10} \)
(ii) \( \int \frac{d x}{4 x^2-4 x+3} \)
(iii) \( \int \frac{dx}{1+x-x^2} \)
Answer:
(i) Let the integral be \( I = \int \frac{d x}{9 x^2+6x+10} \)
First, we factor out 9 from the denominator to make the coefficient of \( x^2 \) equal to 1.
\( I = \frac{1}{9}\int \frac{d x}{x^2+\frac{6}{9} x+\frac{10}{9}} = \frac{1}{9}\int \frac{d x}{x^2+\frac{2}{3} x+\frac{10}{9}} \)
Now, complete the square in the denominator: \( x^2+\frac{2}{3} x+\frac{10}{9} = x^2+\frac{2}{3} x+\left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 + \frac{10}{9} \)
\( = \left(x+\frac{1}{3}\right)^2 - \frac{1}{9} + \frac{10}{9} = \left(x+\frac{1}{3}\right)^2 + \frac{9}{9} = \left(x+\frac{1}{3}\right)^2 + 1^2 \)
So, \( I = \frac{1}{9}\int \frac{d x}{\left(x+\frac{1}{3}\right)^2+1^2} \)
Using the formula \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C \), with \( x = x+\frac{1}{3} \) and \( a = 1 \):
\( I = \frac{1}{9} \times \frac{1}{1}\tan^{-1}\left(\frac{x+\frac{1}{3}}{1}\right) + C \)
\( I = \frac{1}{9}\tan^{-1}\left(\frac{3x+1}{3}\right) + C \)

(ii) Let the integral be \( I = \int \frac{d x}{4 x^2-4 x+3} \)
Factor out 4 from the denominator:
\( I = \frac{1}{4}\int \frac{d x}{x^2-x+\frac{3}{4}} \)
Complete the square in the denominator: \( x^2-x+\frac{3}{4} = x^2-x+\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + \frac{3}{4} \)
\( = \left(x-\frac{1}{2}\right)^2 - \frac{1}{4} + \frac{3}{4} = \left(x-\frac{1}{2}\right)^2 + \frac{2}{4} = \left(x-\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 \)
So, \( I = \frac{1}{4}\int \frac{d x}{\left(x-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2} \)
Using the formula \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C \):
\( I = \frac{1}{4} \times \frac{1}{\frac{1}{\sqrt{2}}}\tan^{-1}\left(\frac{x-\frac{1}{2}}{\frac{1}{\sqrt{2}}}\right)+C \)
\( I = \frac{1}{4} \times \sqrt{2}\tan^{-1}\left(\frac{2x-1}{\sqrt{2}}\right)+C \)
\( I = \frac{\sqrt{2}}{4}\tan^{-1}\left(\frac{2x-1}{\sqrt{2}}\right)+C \)
\( I = \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{2x-1}{\sqrt{2}}\right)+C \)

(iii) Let the integral be \( I = \int \frac{dx}{1+x-x^2} \)
We can rewrite the denominator by taking out a minus sign:
\( I = \int \frac{dx}{-(x^2-x-1)} = -\int \frac{dx}{x^2-x-1} \)
Complete the square in the denominator: \( x^2-x-1 = x^2-x+\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 - 1 \)
\( = \left(x-\frac{1}{2}\right)^2 - \frac{1}{4} - 1 = \left(x-\frac{1}{2}\right)^2 - \frac{5}{4} = \left(x-\frac{1}{2}\right)^2 - \left(\frac{\sqrt{5}}{2}\right)^2 \)
So, \( I = -\int \frac{d x}{\left(x-\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2} \)
Using the formula \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C \), with \( x = x-\frac{1}{2} \) and \( a = \frac{\sqrt{5}}{2} \):
\( I = -\frac{1}{2 \times \frac{\sqrt{5}}{2}} \log\left|\frac{\left(x-\frac{1}{2}\right)-\frac{\sqrt{5}}{2}}{\left(x-\frac{1}{2}\right)+\frac{\sqrt{5}}{2}}\right|+C \)
\( I = -\frac{1}{\sqrt{5}}\log\left|\frac{2x-1-\sqrt{5}}{2x-1+\sqrt{5}}\right|+C \)
In simple words: For each integral, we first adjusted the expression inside by completing the square in the bottom part. This helps to change the original expression into a form that matches known integration rules. Then, we apply the correct formula to find the integral, making sure to handle any negative signs carefully.

🎯 Exam Tip: When the coefficient of \( x^2 \) is not 1, factor it out first. If the quadratic term is negative, factor out the negative sign and then complete the square.

Question 3. Integrate the expression: \( \int \frac{d x}{x\left(x^6+1\right)} \)
Answer:
Let the integral be \( I = \int \frac{d x}{x\left(x^6+1\right)} \)
We can multiply the numerator and denominator by \( x^5 \):
\( I = \int \frac{x^5 d x}{x^6\left(x^6+1\right)} \)
Now, let's use substitution. Put \( x^6 = t \).
Then, differentiate both sides: \( 6x^5 dx = dt \), which means \( x^5 dx = \frac{1}{6} dt \).
Substitute these into the integral:
\( I = \int \frac{\frac{1}{6} dt}{t(t+1)} = \frac{1}{6}\int \frac{d t}{t(t+1)} \)
We can write \( \frac{1}{t(t+1)} \) as \( \frac{1}{t} - \frac{1}{t+1} \) using partial fractions, but the solution shows completing the square for the denominator \( t(t+1) \). Let's follow that path:
\( t(t+1) = t^2+t = t^2+t+\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \left(t+\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \)
So, \( I = \frac{1}{6}\int \frac{d t}{\left(t+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
Using the formula \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C \), with \( x = t+\frac{1}{2} \) and \( a = \frac{1}{2} \):
\( I = \frac{1}{6} \times \frac{1}{2 \times \frac{1}{2}} \log \left|\frac{\left(t+\frac{1}{2}\right)-\frac{1}{2}}{\left(t+\frac{1}{2}\right)+\frac{1}{2}}\right|+C \)
\( I = \frac{1}{6} \times \frac{1}{1} \log \left|\frac{t}{t+1}\right|+C \)
Substitute back \( t = x^6 \):
\( I = \frac{1}{6}\log\left|\frac{x^6}{x^6+1}\right|+C \)
In simple words: To solve this, we first multiplied the top and bottom of the fraction by \( x^5 \) to make it easier to substitute. Then, we let \( x^6 \) be a new variable, \( t \). After changing the integral to be in terms of \( t \), we used a known integration rule for fractions with squared terms to find the answer. Finally, we put \( x^6 \) back in place of \( t \).

🎯 Exam Tip: For integrals like this where a term and its power are involved, try a substitution that makes the higher power the new variable. Remember to adjust the \( dx \) term accordingly.

Question 4. Integrate the expression: \( \int \frac{5 x-2}{1+2 x+3 x^2} d x \)
Answer:
Let the integral be \( I = \int \frac{5 x-2}{3 x^2+2 x+1} d x \)
For integrals of the form \( \int \frac{Px+Q}{Ax^2+Bx+C} dx \), we write \( Px+Q = A'(2Ax+B) + B' \).
Here, \( Px+Q = 5x-2 \) and \( Ax^2+Bx+C = 3x^2+2x+1 \). So, \( 2Ax+B = 6x+2 \).
We need to express \( 5x-2 \) as \( A'(6x+2) + B' \).
Comparing coefficients: \( 6A' = 5 \implies A' = \frac{5}{6} \).
Also, \( 2A' + B' = -2 \). So, \( 2\left(\frac{5}{6}\right) + B' = -2 \implies \frac{5}{3} + B' = -2 \implies B' = -2 - \frac{5}{3} = -\frac{11}{3} \).
So, \( 5x-2 = \frac{5}{6}(6x+2) - \frac{11}{3} \).
Now substitute this back into the integral:
\( I = \int \frac{\frac{5}{6}(6x+2) - \frac{11}{3}}{3 x^2+2 x+1} d x \)
Split the integral into two parts:
\( I = \frac{5}{6}\int \frac{6x+2}{3 x^2+2 x+1} d x - \frac{11}{3}\int \frac{1}{3 x^2+2 x+1} d x \)
For the first part, let \( t = 3x^2+2x+1 \). Then \( dt = (6x+2)dx \).
So, the first integral becomes \( \frac{5}{6}\int \frac{dt}{t} = \frac{5}{6}\log|t| = \frac{5}{6}\log|3x^2+2x+1| \).
For the second part, \( I_2 = \int \frac{1}{3 x^2+2 x+1} d x \). Factor out 3 from the denominator:
\( I_2 = \frac{1}{3}\int \frac{1}{x^2+\frac{2}{3} x+\frac{1}{3}} d x \)
Complete the square in the denominator: \( x^2+\frac{2}{3} x+\frac{1}{3} = x^2+\frac{2}{3} x+\left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 + \frac{1}{3} \)
\( = \left(x+\frac{1}{3}\right)^2 - \frac{1}{9} + \frac{3}{9} = \left(x+\frac{1}{3}\right)^2 + \frac{2}{9} = \left(x+\frac{1}{3}\right)^2 + \left(\frac{\sqrt{2}}{3}\right)^2 \)
So, \( I_2 = \frac{1}{3}\int \frac{1}{\left(x+\frac{1}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2} d x \)
Using the formula \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C \):
\( I_2 = \frac{1}{3} \times \frac{1}{\frac{\sqrt{2}}{3}}\tan^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right) + C \)
\( I_2 = \frac{1}{3} \times \frac{3}{\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) + C \)
\( I_2 = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) + C \)
Now combine both parts:
\( I = \frac{5}{6}\log|3x^2+2x+1| - \frac{11}{3} \times \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) + C \)
\( I = \frac{5}{6}\log|3x^2+2x+1| - \frac{11}{3\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) + C \)
In simple words: When the top of the fraction has \( x \) and the bottom has \( x^2 \), we split the problem into two parts. One part becomes a logarithm after a simple substitution. The other part needs completing the square in the denominator and then using the tangent inverse formula. We make sure to combine both results at the end.

🎯 Exam Tip: Always remember the method for integrals of the form \( \int \frac{Px+Q}{Ax^2+Bx+C} dx \), where you express the numerator in terms of the derivative of the denominator.

Question 5. Evaluate the following integrals.
(ii) \( \int \frac{e^x}{x^{2 x}+6 e^x+5} d x \)
(iii) \( \int \frac{\cos x}{(1-\sin x)(2-\sin x)} dx \)
(iv) \( \int \frac{d x}{x\left[10+7 \log x+(\log x)^2\right]} \)
(v) \( \int \frac{(3 \sin \theta-2) \cos \theta}{5-\cos ^2 \theta-4 \sin \theta} d \theta \)
Answer:
(i) Let the integral be \( I = \int \frac{x}{x^4+2x^2+3} d x \)
We use substitution here. Put \( x^2 = t \).
Differentiate both sides: \( 2xdx = dt \), so \( xdx = \frac{1}{2} dt \).
Substitute into the integral:
\( I = \int \frac{\frac{1}{2} dt}{t^2+2t+3} = \frac{1}{2}\int \frac{d t}{t^2+2t+3} \)
Complete the square in the denominator: \( t^2+2t+3 = t^2+2t+1+2 = (t+1)^2+(\sqrt{2})^2 \).
So, \( I = \frac{1}{2}\int \frac{d t}{(t+1)^2+(\sqrt{2})^2} \)
Using the formula \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C \):
\( I = \frac{1}{2} \times \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t+1}{\sqrt{2}}\right) + C \)
Substitute back \( t = x^2 \):
\( I = \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x^2+1}{\sqrt{2}}\right) + C \)

(ii) Let the integral be \( I = \int \frac{e^x}{e^{2 x}+6 e^x+5} d x \)
Use substitution: Put \( e^x = t \).
Differentiate both sides: \( e^x dx = dt \).
Substitute into the integral:
\( I = \int \frac{d t}{t^2+6t+5} \)
Complete the square in the denominator: \( t^2+6t+5 = t^2+6t+9-4 = (t+3)^2-2^2 \).
So, \( I = \int \frac{d t}{(t+3)^2-2^2} \)
Using the formula \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C \):
\( I = \frac{1}{2 \times 2}\log\left|\frac{(t+3)-2}{(t+3)+2}\right|+C \)
\( I = \frac{1}{4}\log\left|\frac{t+1}{t+5}\right|+C \)
Substitute back \( t = e^x \):
\( I = \frac{1}{4}\log\left|\frac{e^x+1}{e^x+5}\right|+C \)

(iii) Let the integral be \( I = \int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x \)
Use substitution: Put \( \sin x = t \).
Differentiate both sides: \( \cos x dx = dt \).
Substitute into the integral:
\( I = \int \frac{d t}{(1-t)(2-t)} \)
We can multiply the terms in the denominator: \( (1-t)(2-t) = 2-t-2t+t^2 = t^2-3t+2 \).
So, \( I = \int \frac{d t}{t^2-3t+2} \)
Complete the square in the denominator: \( t^2-3t+2 = t^2-3t+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+2 \)
\( = \left(t-\frac{3}{2}\right)^2-\frac{9}{4}+2 = \left(t-\frac{3}{2}\right)^2-\frac{1}{4} = \left(t-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2 \)
So, \( I = \int \frac{d t}{\left(t-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
Using the formula \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C \):
\( I = \frac{1}{2 \times \frac{1}{2}}\log\left|\frac{\left(t-\frac{3}{2}\right)-\frac{1}{2}}{\left(t-\frac{3}{2}\right)+\frac{1}{2}}\right|+C \)
\( I = \log\left|\frac{t-2}{t-1}\right|+C \)
Substitute back \( t = \sin x \):
\( I = \log\left|\frac{\sin x-2}{\sin x-1}\right|+C \)

(iv) Let the integral be \( I = \int \frac{d x}{x\left[10+7 \log x+(\log x)^2\right]} \)
Use substitution: Put \( \log x = t \).
Differentiate both sides: \( \frac{1}{x} dx = dt \).
Substitute into the integral:
\( I = \int \frac{d t}{t^2+7t+10} \)
Complete the square in the denominator: \( t^2+7t+10 = t^2+7t+\left(\frac{7}{2}\right)^2-\left(\frac{7}{2}\right)^2+10 \)
\( = \left(t+\frac{7}{2}\right)^2-\frac{49}{4}+10 = \left(t+\frac{7}{2}\right)^2-\frac{9}{4} = \left(t+\frac{7}{2}\right)^2-\left(\frac{3}{2}\right)^2 \)
So, \( I = \int \frac{d t}{\left(t+\frac{7}{2}\right)^2-\left(\frac{3}{2}\right)^2} \)
Using the formula \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C \):
\( I = \frac{1}{2 \times \frac{3}{2}}\log\left|\frac{\left(t+\frac{7}{2}\right)-\frac{3}{2}}{\left(t+\frac{7}{2}\right)+\frac{3}{2}}\right|+C \)
\( I = \frac{1}{3}\log\left|\frac{t+2}{t+5}\right|+C \)
Substitute back \( t = \log x \):
\( I = \frac{1}{3}\log\left|\frac{\log x+2}{\log x+5}\right|+C \)

(v) Let the integral be \( I = \int \frac{(3 \sin \theta-2) \cos \theta}{5-\cos ^2 \theta-4 \sin \theta} d \theta \)
First, we use the identity \( \cos^2 \theta = 1-\sin^2 \theta \) in the denominator.
Denominator: \( 5-(1-\sin^2 \theta)-4 \sin \theta = 5-1+\sin^2 \theta-4 \sin \theta = \sin^2 \theta-4 \sin \theta+4 \).
So, \( I = \int \frac{(3 \sin \theta-2) \cos \theta}{\sin ^2 \theta-4 \sin \theta+4} d \theta \)
Use substitution: Put \( \sin \theta = t \).
Differentiate both sides: \( \cos \theta d \theta = dt \).
Substitute into the integral:
\( I = \int \frac{3t-2}{t^2-4t+4} dt \)
Notice that the denominator is a perfect square: \( t^2-4t+4 = (t-2)^2 \).
So, \( I = \int \frac{3t-2}{(t-2)^2} dt \)
We can rewrite the numerator \( 3t-2 \) in terms of \( (t-2) \):
\( 3t-2 = 3(t-2) + 6 - 2 = 3(t-2) + 4 \).
So, \( I = \int \frac{3(t-2)+4}{(t-2)^2} dt \)
Split the integral into two parts:
\( I = \int \left(\frac{3(t-2)}{(t-2)^2} + \frac{4}{(t-2)^2}\right) dt \)
\( I = \int \frac{3}{t-2} dt + \int \frac{4}{(t-2)^2} dt \)
\( I = 3\int \frac{1}{t-2} dt + 4\int (t-2)^{-2} dt \)
Integrate each part:
\( I = 3\log|t-2| + 4 \frac{(t-2)^{-2+1}}{-2+1} + C \)
\( I = 3\log|t-2| + 4 \frac{(t-2)^{-1}}{-1} + C \)
\( I = 3\log|t-2| - \frac{4}{t-2} + C \)
Substitute back \( t = \sin \theta \):
\( I = 3\log|\sin \theta-2| - \frac{4}{\sin \theta-2} + C \)
In simple words: These problems often need a smart substitution to change them into simpler forms. We look for parts of the integral that, when replaced by a new variable, make the whole expression easier to handle. Sometimes, this means completing the square or breaking the numerator into parts to match standard formulas. Then, we solve and substitute the original variable back.

🎯 Exam Tip: For substitution problems, choose a variable \(t\) such that its derivative is also present in the integral. For trigonometric integrals, identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) are often key to simplifying the denominator.