$\int\frac{1}{x^2(x^4+1)^{3/4}}dx$ is equal to:

$(1+\frac{1}{x^4})^{1/4}+C$ $(x^4+1)^{1/4}+C$ $(1-\frac{1}{x^4})^{1/4}+C$ $-(1+\frac{1}{x^4})^{1/4}+C$

$-(1+\frac{1}{x^4})^{1/4}+C$

$\int\frac{1}{x^2(x^4+1)^{3/4}}dx=\int\frac{1}{x^5(1+\frac{1}{x^4})^{3/4}}dx$ $=\frac{-1}{4}\int\frac{1}{t^{3/4}}dt=-\frac{1}{4}\frac{t^{1/4}}{1/4}+C=-t^{1/4}+C$, where $t=\frac{1}{1+\frac{1}{x^4}}=-(1+\frac{1}{x^4})^{1/4}+C$