Question

$\int \frac{x^{3}dx}{1+x^8}=$

• 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

Answer: (B)

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