Question

$\int \frac{x^2}{1+x^6}dx$ is equal to

• A. $ta{n}^{-1}{x}^3+c$
• B. $ta{n}^{-1}{x}^2+c$
• C. $\frac{1}{3}ta{n}^{-1}{x}^3+c$
• D. $3ta{n}^{-1}{x}^3+c$

Answer 1

The correct option is C $\frac{1}{3}ta{n}^{-1}{x}^3+c$

$\int \frac{x^2}{1+x^6}dx=\frac{1}{3}\int \frac{3x^2}{1+(x^3{)}^2}dx$

Now, Let $t=x^{3}dt=3x^{2}dx$

$\frac{1}{3}\int \frac{dt}{1+(t{)}^2}=\frac{1}{3}{\tan }^{-1}t+c=\frac{1}{3}{\tan }^{-1}{x}^3+c$