Question

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

• A. $\frac{1}{2}{\tan }^{-1}{x}^3.$
• B. ${\tan }^{-1}{x}^3.$
• C. ${\tan }^{-1}{x}^6.$
• D. ${\sec }^{-1}{x}^3.$

Answer 1

Answer: (B)

${\tan }^{-1}{x}^3.$

Let $I=\int \frac{3x^2}{x^6+1}dx$
Put $\sqrt{x^3}=t\Rightarrow \frac{3\sqrt{x}}{2}dx=dt$
Therefore
$I=2\int \frac{dt}{1+t^2}={\tan }^{-1}t={\tan }^{-1}{x}^3$
Hence, option 'B' is correct.