Question

$\int \frac{x^{3}dx}{1+x^8}$ is equal to

• A. $4{\tan }^{-1}{x}^3+C$
• B. $\frac{1}{4}{\tan }^{-1}{x}^4+C$
• C. $x+4{\tan }^{-1}{x}^4+C$
• D. $x^2+\frac{1}{4}{\tan }^{-1}{x}^4+C$

Answer 1

The correct option is B $\frac{1}{4}{\tan }^{-1}{x}^4+C$

Let $I=\int \frac{x^{3}dx}{1+x^8}$
Put $x^4=t\Rightarrow 4x^{3}dx=dt$
$\therefore I=\int \frac{dt}{4(1+t^2)}$
$=\frac{1}{4}{\tan }^{-1}t+C$

$=\frac{1}{4}{\tan }^{-1}{x}^4+C$