Question

$\int \frac{x^{2}}{1+\left(x^{3}\right)^{2}}dx $ is equal to

• A. $\tan^{-1} x^2 + C $
• B. $2/3 \tan^{-1} x^3 + C $
• C. $1/3 \tan^{-1} (x^3) + C $
• D. $1/2 \tan^{-1} x^2 + C $
• E. $\tan^{-1} x^3 + C $

Answer 1

Answer: (C)

Let $I=\int \frac{x^{2}}{1+\left(x^{3}\right)^{2}} d x$
Put $x^{3}=t$
$\therefore 3 x^{2} d x=d t$
$\therefore I=\frac{1}{3} \int \frac{d t}{1+t^{2}}=\frac{1}{3} \tan ^{-1} t+C$
$=\frac{1}{3} \tan ^{-1}\left(x^{3}\right)+C$