$$\int \frac{1}{\sqrt{1+x^2}(1+x^2)} dx$$
I tried using substitution because I know that $$\frac{1}{(1+x^2)}$$ is the derivation of arcta(x).
But I get stuck since puting u= arctan(x) does not help me.. I also rewrote it as $$\int \frac{1}{(1+x^2)^{\frac{3}{2}}} dx$$ But I still can't get it..
Let $x=\tan(u)$ so $dx=\sec^2(u)\ du$, and don't forget $1+\tan^2=\sec^2$.
$$\begin{align}\int\frac1{(1+x^2)^{3/2}}\ dx&=\int\frac{\sec^2(u)}{(1+\tan^2(u))^{3/2}}\ du\\&=\int\frac{\sec^2(u)}{\sec^3(u)}\ du\\&=\int\cos(u)\ du\\&=\sin(u)+c\\&=\sin(\arctan(x))+c\\&=\frac x{\sqrt{1+x^2}}+c\end{align}$$