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

To integrate $I = \int \frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}} dx$,we manipulate the integrand.
Multiply and divide by $x^{-3}$:
$I = \int \frac{x^{-3}}{x^{2} x^{-3}\left(x^{4}+1\right)^{\frac{3}{4}}} dx = \int \frac{x^{-3}}{\left(x^{4}+1\right)^{\frac{3}{4}} x^{-1}} dx$
Rewrite the expression inside the integral:
$I = \int \frac{x^{-3}}{\left(x^{4}\left(1+\frac{1}{x^{4}}\right)\right)^{\frac{3}{4}}} dx = \int \frac{x^{-3}}{x^{4 \cdot \frac{3}{4}} \left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}} dx$
$I = \int \frac{x^{-3}}{x^{3} \left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}} dx = \int \frac{1}{x^{6}} \left(1+\frac{1}{x^{4}}\right)^{-\frac{3}{4}} dx$
Let $t = 1 + \frac{1}{x^{4}}$. Then $dt = -\frac{4}{x^{5}} dx$,which implies $\frac{1}{x^{5}} dx = -\frac{dt}{4}$.
Wait,let's re-evaluate the substitution:
$I = \int \frac{1}{x^{2} \cdot x^{3} (1 + x^{-4})^{3/4}} dx = \int x^{-5} (1 + x^{-4})^{-3/4} dx$
Let $u = 1 + x^{-4}$. Then $du = -4x^{-5} dx$,so $x^{-5} dx = -\frac{du}{4}$.
$I = -\frac{1}{4} \int u^{-3/4} du = -\frac{1}{4} \left( \frac{u^{1/4}}{1/4} \right) + C = -u^{1/4} + C$
Substituting back $u = 1 + \frac{1}{x^{4}}$:
$I = -\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}} + C$