How to use U substitution for the integral $\int\frac{8}{49+x^2}\,dx$? So, the following is my problem. 
$$\int\frac{8}{49+x^2}\,dx$$
I understand this. I should first take out the constant which is 8 so it'll be
$$8\int\frac{1}{49+x^2}\,dx$$
Then I should factor out the 49 no? So it'll be
$$\frac{8}{49}\int\frac{1}{1+\frac{x^2}{49}}\,dx$$
Let $u=\dfrac{x}{7}$ and thus $du=\dfrac{1}{7}\,dx$. I'm stuck at this point.

Here are the steps 
$$\int\frac{8}{x^2+49}dx= 8\int\frac{1}{x^2+7^2}dx 
$$
$$=\frac{8}{7^2} \int\frac{1}{\frac{x^2}{7^2}+1}dx =\frac{8}{7^2} \int\frac{1}{\left(\frac{x}{7}\right)^2+1}dx $$
Let $u=\frac{x}{7}$, then $du=\frac{1}{7}dx$. So now
$$ \frac{8}{7} \int\frac{1}{u^2+1}du= \frac{8}{7} \arctan u+C $$
$$ = \frac{8}{7} \arctan\left(\frac{x}{7}\right)+C $$
