Question:
Evaluate the following integrals:
$\int \frac{1}{x^{2}\left(x^{4}+1\right)^{3 / 4}} d x$
Solution:
$I=\int \frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}} d x$
$\Rightarrow \int \frac{1}{x^{5}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}} d x$
Let $1+\frac{1}{x^{4}}=t$
$\Rightarrow-\frac{4}{x^{5}} d x=d t$
$\Rightarrow \frac{1}{x^{5}} d x=\frac{-d t}{4}$
$I=\frac{-1}{4} \int \frac{1}{t^{\frac{3}{4}}} d t$
$\Rightarrow \frac{-1}{4}\left(\frac{\mathrm{t}^{\frac{1}{4}}}{\frac{1}{4}}\right)+\mathrm{C}$
$\Rightarrow-t^{\frac{1}{4}}+c$
But $t=1+\frac{1}{x^{4}}$
$\Rightarrow-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c$
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