Show that

$\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}$

Multiplying and dividing by $x^{-3}$, we obtain

$\frac{x^{-3}}{x^{2} \cdot x^{-3}\left(x^{4}+1\right)^{\frac{3}{4}}}=\frac{x^{-3}\left(x^{4}+1\right)^{\frac{-3}{4}}}{x^{2} \cdot x^{-3}}$

$=\frac{\left(x^{4}+1\right)^{\frac{-3}{4}}}{x^{5} \cdot\left(x^{4}\right)^{-\frac{3}{4}}}$

$=\frac{1}{x^{5}}\left(\frac{x^{4}+1}{x^{4}}\right)^{-\frac{3}{4}}$

$=\frac{1}{x^{5}}\left(1+\frac{1}{x^{4}}\right)^{-\frac{3}{4}}$

Let $\frac{1}{x^{4}}=t \Rightarrow-\frac{4}{x^{5}} d x=d t \Rightarrow \frac{1}{x^{5}} d x=-\frac{d t}{4}$

$\therefore \int \frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}} d x=\int \frac{1}{x^{5}}\left(1+\frac{1}{x^{4}}\right)^{-\frac{3}{4}} d x$

$=-\frac{1}{4} \int(1+t)^{-\frac{3}{4}} d t$

$=-\frac{1}{4}\left[\frac{(1+t)^{\frac{1}{4}}}{\frac{1}{4}}\right]+\mathrm{C}$

$=-\frac{1}{4} \frac{\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}}{\frac{1}{4}}+\mathrm{C}$

$=-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+\mathrm{C}$