# Evaluate $\int\frac{1}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}dx$

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?

• I would recommend starting with a change of variable, $x=u^{25}$. – Barry Cipra Jul 18 at 19:03
• And then integration by parts. – Viktor Glombik Jul 18 at 19:11

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 $$x=y^{25}\implies dx=25\,y^{24}\,dy$$ $$\int\frac{dx}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}=25\int\frac{ y^{15}}{y^{32}+1}\,dy=\frac {25}{16}\int\frac{16\, y^{15}}{y^{32}+1}\,dy=\frac {25}{16}\int\frac{ \left(y^{16}\right)'}{\left(y^{16}\right)^2+1}$$