The integral $\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x$ is equal to (where $C$ is a constant of integration)

(B) $\frac{x^{12}}{6\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C$

(A) $\frac{x^{4}}{6\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C$

(C) $\frac{x^{4}}{\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C$

(D) $\frac{x^{12}}{\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C$

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Indefinite Integrals

The integral

\int \frac{3 x ^{13} + 2 x ^{11}}{( 2 x ^{4} + 3 x ^{2} + 1 ) ^{4}} d x

is equal to (where

C

is a constant of integration)

Answer: (B)

Divide numerator and denominator by

x^{16}

to get

\int \frac{( \frac{3}{x ^{3}} + \frac{2}{x ^{5}} ) d x}{( 2 + \frac{3}{x ^{2}} + \frac{1}{x ^{4}} ) ^{4}}

. With

t = 2 + \frac{3}{x ^{2}} + \frac{1}{x ^{4}}

d t = - 2 (\frac{3}{x ^{3}} + \frac{2}{x ^{5}}) d x

, so the integral becomes

- \frac{1}{2} \int t^{- 4} d t = \frac{1}{6 t ^{3}} + C = \frac{x ^{12}}{6 ( 2 x ^{4} + 3 x ^{2} + 1 ) ^{3}} + C

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option