Calculate$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$

Hint: Let $y = (x+2)e^{-x/2}.$ You should get $I=-2\sinh^{-1}{2}$.

$\displaystyle y = (x+2)e^{-x/2} \implies dy = -\frac{x}{2}e^{-x/2}\,dx$ and $\displaystyle \frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = \frac{xe^{-x/2}\,dx}{\sqrt{1+(x+2)^2e^{-x}}} = \frac{-2\,dy}{\sqrt{1+y^2}}.$

Therefore we have $\displaystyle I = \int_{-2}^{0}\frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = -2\int_{0}^{2}\frac{dy}{\sqrt{1+y^2}} = -2\sinh^{-1}(2)$, as claimed.

Tags:

Integration

Calculus

Definite Integrals

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