Question
Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}^2+4)\sqrt{\text{x}^2+9}}\text{ dx}$

Answer

$\text{I}=\int\frac{\text{x}}{(\text{x}^2+4)\sqrt{\text{x}^2+9}}\text{ dx}$
Let $\text{x}^2+9=\text{u}^2$
$2\text{xdx}=2\text{udu}$
$\therefore\ \text{I}=\int\frac{\text{u}}{(\text{u}^2-5)\text{u}}\text{ du}$
$=\int\frac{\text{du}}{\text{u}^2-5}$
$=\frac{1}{2\sqrt{5}}\log\Big(\frac{\text{u}-\sqrt{5}}{\text{u}+\sqrt{5}}\Big)+\text{C}$
$=\frac{1}{2\sqrt{5}}\log\bigg(\frac{\sqrt{\text{x}^2+9}-\sqrt{5}}{\sqrt{\text{x}^2+9}+\sqrt{5}}\bigg)+\text{C}$

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