Compute$\int\limits_{0}^{2} \sqrt{x^2-2x+2}\ln(2+x)dx$.

Question

Compute: $I=\displaystyle \int\limits_{0}^{2} \sqrt{x^2-2x+2}\ln(2+x)dx$.

I tried to : $I=\displaystyle \int \limits_{-1}^{1}\sqrt{t^2+1}\ln(3+t)dt$ set $t=\tan u\Rightarrow dt=(1+\tan^2u)du$

and $I=\displaystyle \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \dfrac{(\ln(4+\tan^2u))du}{\cos^3u}$

Answer 1

Hint:

1. ${x+2=e^z \implies dx=e^zdz}$

$\displaystyle \int \sqrt{x^2-2x+2}\ln(2+x)dx\\=\displaystyle \int ze^z\sqrt{e^{2z}-6e^z+10}\ dz\\=z\displaystyle \int e^z\sqrt{e^{2z}-6e^z+10}\ dz-\displaystyle \int \left(\displaystyle\int e^z\sqrt{e^{2z}-6e^z+10}\ dz\right)dz$

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2. $e^z=y \implies e^zdz = dy$

$\therefore\displaystyle\int e^z\sqrt{e^{2z}-6e^z+10}\ dz=\displaystyle\int \sqrt{y^2-6y^2+10}\ dy=\displaystyle\int \sqrt{(y-3)^2+1}\ dy$

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3. $y-3=t \implies dy=dt$

$\therefore \displaystyle\int \sqrt{(y-3)^2+1}\ dy=\displaystyle\int \sqrt{t^2+1}\ dt=\dfrac{t\sqrt{t^2+1}}{2}+\dfrac{1}{2}\operatorname{arsinh}t$

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4. $\displaystyle\int\operatorname{arsinh}t\ dt=t \operatorname{arsinh}t -{\sqrt{t^2+1}}+C$