Question

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

Answer 1

$$\begin{gathered} \int \frac{dx}{x^2{\left(x^4+1\right)}^{{\displaystyle \frac{3}{4}}}} \\ =\int \frac{dx}{x^2{\left[x^4\left(1+{\displaystyle \frac{1}{x^4}}\right)\right]}^{{\displaystyle \frac{3}{4}}}} \\ =\int \frac{dx}{x^2.x^3{\left(1+{\displaystyle \frac{1}{x^4}}\right)}^{{\displaystyle \frac{3}{4}}}} \\ =\int \frac{{\left(1+{\displaystyle \frac{1}{x^4}}\right)}^{-{\displaystyle \frac{3}{4}}}}{x^5}dx \\ \mathrm{Let}1+\frac{1}{x^4}=t \\ \Rightarrow -\frac{4}{x^5}dx=dt \\ \Rightarrow \frac{dx}{x^5}=-\frac{dt}{4} \\ \mathrm{Now},\int \frac{{\left(1+{\displaystyle \frac{1}{x^4}}\right)}^{-{\displaystyle \frac{3}{4}}}}{x^5}dx \\ =-\frac{1}{4}\int t^{-\frac{3}{4}}dt \\ =-\frac{1}{4}\left[\frac{t^{-{\displaystyle \frac{3}{4}}+1}}{-{\displaystyle \frac{3}{4}}+1}\right]+C \\ =-t^{\frac{1}{4}}+C \\ =-{\left(1+\frac{1}{x^4}\right)}^{\frac{1}{4}}+C \end{gathered}$$