How to integrate $\frac{dx}{ \left(1+x^2 \right)^2}$ and$\frac{\sqrt[6]{x}}{1+\sqrt[3]{x}}$

How to integrate $\int \frac{dx}{ \left(1+x^2 \right)^2}$ and$\int \frac{\sqrt[6]{x}}{1+\sqrt[3]{x}} dx$ ? Which method (by parts or substitution )should be applied? How to use this method in those examples?

• Set $x=\tan t$ in the first integral and $x=u^6$ in the second one. After that, everything is pretty easy. – Jack D'Aurizio Apr 11 '15 at 19:44
• I do not get it, how the first substitution helps, could you elaborate on that further? – Krowskir Apr 11 '15 at 19:48

The first can be done by parts: write it as $$\int \frac{dx}{(1+x^2)^2} = \int \frac{dx}{1+x^2} - \int \frac{x^2}{(1+x^2)^2} \, dx,$$ and then $$\int \frac{x^2}{(1+x^2)^2} \, dx = -\frac{1}{2}\frac{x}{1+x^2} + \frac{1}{2}\int \frac{dx}{1+x^2},$$ and shoving this into the previous equation gives the answer as $$\frac{1}{2}\frac{x}{1+x^2} +\frac{1}{2} \arctan{x}+C.$$
The second is conventionally done with $x=u^6$, so $dx = 6u^5 \, du$, and we have to do $$6\int \frac{u^6}{1+u^2} \, du,$$ which is then just a partial fractions calculation.
By setting $x=\tan t$ in the first integral we have $dx=\frac{dt}{\cos^2 t}$, from which: $$\int \frac{dx}{(1+x^2)^2} = \int \cos^2(t)\,dt = \frac{t+\sin t\cos t}{2}=\frac{1}{2}\left(\arctan x+\frac{x}{1+x^2}\right).$$ By setting $x=u^6$ in the second integral we have $dx = 6 u^5\,du$, from which: $$\int\frac{\sqrt[6]{x}}{1+\sqrt[3]{x}}\,dx = 6\int \frac{u^6}{u^2+1}\,du = 6\left(\int (u^4-u^2+1)\,du-\int\frac{du}{u^2+1}\right).$$