Evaluate the following integrals: 1 (x^2+4) x^2+1 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{(\text{x}^2+4)\sqrt{\text{x}^2+1}}\text{ dx}$

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Answer

$\text{I}=\int\frac{1}{(\text{x}^2+4)\sqrt{\text{x}^2+1}}\text{ dx}$
Let $\text{x}^2+1=\text{u}^2$
$2\text{xdx}=2\text{udu}$
$\therefore\ \text{I}=\int\frac{\text{u}}{(\text{u}^2+3)\text{u}}\text{ du}$
$=\int\frac{1}{\text{u}^2+3}\text{ du}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{u}}{\sqrt{3}}\Big)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\bigg(\sqrt{\frac{\text{x}^2+1}{3}}\bigg)+\text{C}$

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