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