Question

$\int \frac{x^3{\tan }^{-1}{x}^4}{1+x^8}dx=$

• A. $\frac{{\tan }^{-1}{x}^4}{4}+c$
• B. $\frac{x^3{\tan }^{-1}{x}^4}{4}+c$
• C. $\frac{1}{8}({\tan }^{-1}{x}^4{)}^2+c$
• D. $\frac{x^2{\tan }^{-1}{x}^4}{8}+c$

Answer 1

The correct option is C $\frac{1}{8}({\tan }^{-1}{x}^4{)}^2+c$

$\int x^3\frac{ta{n}^{-1}{x}^4}{1+x^8}\cdot dx$
$x^4=t$
$\Rightarrow 4x^3\cdot dx=dt$
$\Rightarrow x^3\cdot dx=\frac{1}{4}\cdot dt$
$\frac{1}{4}\int \frac{ta{n}^{-1}t\cdot dt}{1+t^2}$
$=\frac{1}{4}\int ta{n}^{-1}t\cdot d\left(ta{n}^{-1}t\right)$
$=\frac{1}{4}[\frac{(ta{n}^{-1}t{)}^2}{2}+c]$
$=\frac{1}{8}(ta{n}^{-1}t{)}^2+c$
$=\frac{1}{8}(ta{n}^{-1}{x}^4{)}^2+c\cdot$