Evaluate $$\int\frac{1}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}dx$$
I start by factoring $$\int\frac{1}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}dx=\int\frac{1}{x^{\frac{9}{25}}\left(x^{\frac{32}{25}}+1\right)}dx$$
Can I do partial fraction from here?
This is a cute $u$-substitution problem. The crux of the problem is dividing out by the right factor of $x$. Notice that if we factor out one copy of $x$ we obtain
$$ \int \frac{1}{x\cdot x^{\frac{16}{25}} + x^{\frac{9}{25}} }dx \;\; =\;\; \int \frac{1}{x \left (x^{\frac{16}{25}} + x^{\frac{-16}{25}} \right )}dx. $$
Now let $u = x^{\frac{16}{25}}$, and thus $du = \frac{16}{25}x^{-\frac{9}{25}}dx$. What this yields is
\begin{eqnarray*} \int \frac{1}{x \left (x^{\frac{16}{25}} + x^{\frac{-16}{25}} \right )}dx & = & \frac{25}{16}\int \frac{x^{\frac{9}{25}}} { x\left (u + \frac{1}{u} \right )} du \\ & = & \frac{25}{16} \int \frac{1}{x^{\frac{16}{25}} \left (u + \frac{1}{u} \right ) }du \\ & = & \frac{25}{16} \int \frac{1}{u\left (u + \frac{1}{u} \right )} du \\ & = & \frac{25}{16} \int \frac{1}{u^2 + 1} du. \end{eqnarray*}
Take $u = \tan \theta$. Final answer should be $\frac{25}{16}\tan^{-1}\left (x^{\frac{16}{25}}\right ) + c$.
Let $$x=y^{25}\implies dx=25\,y^{24}\,dy$$ $$\int\frac{dx}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}=25\int\frac{ y^{15}}{y^{32}+1}\,dy=\frac {25}{16}\int\frac{16\, y^{15}}{y^{32}+1}\,dy=\frac {25}{16}\int\frac{ \left(y^{16}\right)'}{\left(y^{16}\right)^2+1}$$
