Evaluate the following integrals: x (x^2+2x+2) x+1 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}^2+2\text{x}+2)\sqrt{\text{x}+1}}\text{ dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}}{(\text{x}^2+2\text{x}+2)\sqrt{\text{x}+1}}\text{ dx}$
Let $\text{x}+1=\text{t}^2$
$\text{dx}=2\text{t dt}$
$=2\int\frac{(\text{t}^2-1)\text{t dt}}{(\text{t}^4+1)\text{t}}$
$=2\int\frac{(\text{t}^2-1)\text{ dt}}{(\text{t}^4+1)}$
$=2\int\frac{\big(1-\frac{1}{\text{t}^2}\big)\text{ dt}}{\text{t}+\frac{1}{\text{t}^2}}$
$=2\int\frac{\big(1-\frac{1}{\text{t}^2}\big)}{\big(\text{t}+\frac{1}{\text{t}}\big)^2-2}$
Let $\text{t}+\frac{1}{\text{t}}=\text{y}$
$\Big(1-\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dy}$
$\therefore\ \text{I}=2\int\frac{\text{dy}}{\text{y}^2-2}$
$=\frac{2}{2\sqrt{2}}\log\bigg|\frac{\text{y}-\sqrt{2}}{\text{y}+\sqrt{2}}\bigg|+\text{C}$
Thus, $\text{I}=\frac{1}{\sqrt{2}}\log\bigg|\frac{\text{t}^2+1-\sqrt{2}\text{t}}{\text{t}^2+1+\sqrt{2}\text{t}}\bigg|+\text{C}$
Hence,
$\text{I}=\frac{1}{\sqrt{2}}\log\begin{vmatrix}\frac{\text{x}+2-\sqrt{2(\text{x}+1)}}{\text{x}+2\sqrt{2(\text{x}+1)}}\end{vmatrix}+\text{C}$

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