Answer
$\pi$
Work Step by Step
$\int\frac{4\,dx}{1+(2x+1)^{2}}=4\int\frac{\,dx}{1+(2x+1)^{2}}$
Substituting $u=2x+1$ so that $dx=\frac{du}{2}$, we have
$4\int\frac{\,dx}{(2x+1)^{2}+1}=4\times\frac{1}{2}\int\frac{du}{u^{2}+1}=2\tan^{-1}(u)+C$
$=2\tan^{-1}(2x+1)+C$
Therefore $\int^{0}_{-1}\int\frac{4\,dx}{1+(2x+1)^{2}}=[2\tan^{-1}(2x+1)]^{0}_{-1}$
$=2\times\tan^{-1}(1)-2\times\tan^{-1}(-1)=2\times\frac{\pi}{4}-(2\times-\frac{\pi}{4})=\pi$
