Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

$\int \frac{m x^{m+2 n-1}-n x^{n-1}}{x^{2 m+2 n}+2 x^{m+n}+1} d x$ is equal to

Options:

$\frac{-x^m}{x^{m+n}+1}+C$

$\frac{x^n}{x^{m+n}+1}+C$

$\frac{-x^n}{x^{m+n}+1}+C$

$\frac{x^m}{x^{m+n}+1}+C$

Correct Answer:

$\frac{-x^n}{x^{m+n}+1}+C$

Explanation:

Let $I=\int \frac{m x^{m+2 n-1}-n x^{n-1}}{x^{2 m+2 n}+2 x^{m+n}+1} d x$. Then

$I=\int \frac{\left(m x^{m+n}-n\right) x^{n-1}}{\left(x^{m+n}+1\right)^2} d x$

$\Rightarrow I=\int \frac{(m+n) x^{m+2 n-1}-n x^{m+2 n-1}-n x^{n-1}}{\left(x^{m+n}+1\right)^2} d x$

$\Rightarrow I =\int \frac{(m+n) x^{m+n-1} \times x^n-n x^{n-1}\left(x^{m+n}+1\right)}{\left(x^{m+n}+1\right)^2} d x$

$\Rightarrow I=\int x^n \times \frac{(m+n) x^{m+n-1}}{\left(x^{m+n}+1\right)^2} d x-\int \frac{n x^{n-1}}{\left(x^{m+n}+1\right)} d x $

$\Rightarrow I=\frac{-x^n}{x^{m+n}+1}-\int n x^{n-1} \times \frac{-1}{\left(x^{m+n}+1\right)} d x-\int \frac{n x^{n-1}}{\left(x^{m+n}+1\right)} d x+C$

$\Rightarrow I=\frac{-x^n}{x^{m+n}+1}+C$