\(\displaystyle \int \frac{dx}{4+x^2}  dx \)

Summary Submit

The derivative of arctan:

$$ \frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2} $$

Rewriting our integral:

$$ \int \frac{dx}{4+x^2} = \int \frac{dx}{4(1+\frac{x^2}{4})} $$ $$ = \frac{1}{4}\int \frac{dx}{1+\left(\frac{x}{2}\right)^2}  dx $$

Using substitution:

$$ u=\frac{x}{2} \hskip{2em} du =\frac{dx}{2} \hskip{2em} 2  du=dx$$ $$ \frac{1}{4}\int \frac{2   du}{1+u^2} =\frac{1}{2}\int \frac{du}{1+u^2}$$ $$ = \frac{1}{2}\tan^{-1}u + C$$ $$ = \boxed{\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)+C} $$