Integral $\int_{-2}^0 \frac{x}{\sqrt{e^x+(x+2)^2}}dx$ I am trying to evaluate $$\int_{-2}^0 \frac{x}{\sqrt{e^x+(x+2)^2}}dx$$ So far I had no succes using trig substitution or integration by parts, also some random substitution like $x=2t$ and moved the exponential to the numerator, but I am stuck. Could you perhaps give me an idea? (this is a college admission problem)

Factor out $e^x$ in the denominator. Once you take the square root, you get $e^{x/2}$ in the denominator. Then, make the substitution:
Let $(x+2)e^{-x/2} = \tan \theta$
$-\dfrac{x}{2}e^{-x/2}dx = \sec^2 \theta d\theta$
At $x=-2$, $\tan \theta = 0$
At $x=0$, $\tan \theta = 2$
So, your integral becomes:
$$\int_{-2}^0 \dfrac{x}{\sqrt{e^x+(x+2)^2}}dx = -2\int_0^{\arctan 2} \sec \theta d\theta = -2\ln(\sqrt{5}+2)$$
