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