Evaluate the following integral: 1 a^2+b^2x^2 dx - Vidyadip

Question

Evaluate the following integral:
$\int\frac{1}{\sqrt{\text{a}^2+\text{b}^2\text{x}^2}}\text{ dx}$

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Answer

$\int\frac{1}{\sqrt{\text{a}^2+\text{b}^2\text{x}^2}}\text{ dx}$
$=\int\frac{\text{dx}}{\sqrt{\text{b}^2\Big({\frac{\text{a}^2}{\text{b}^2}+\text{x}^2\Big)}}}$
$=\frac{1}{\text{b}}\int\frac{\text{dx}}{\sqrt{\text{x}^2+\big(\frac{\text{a}}{\text{b}}\big)^2}}$
$=\frac{1}{\text{b}}\log\Big|\text{x}+\sqrt{\text{x}^2+\frac{\text{a}^2}{\text{b}^2}}\Big|+\text{C}$
$=\frac{1}{\text{b}}\Big[\log\Big|\text{x}+\frac{\sqrt{\text{b}^2\text{x}^2+\text{a}^2}}{\text{b}}\Big|\Big]+\text{C}$
$=\frac{1}{\text{b}}\Big[\log\Big|\frac{\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}}{\text{b}}\Big|\Big]+\text{C}$
$=\frac{1}{\text{b}}\Big[\log\big|\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}\big|-\log\text{b}\Big]+\text{C}$
$=\frac{1}{\text{b}}\log\big|\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}\big|-\frac{\log\text{b}}{\text{b}}+\text{C}$
Let $\text{C}-\frac{\log\text{b}}{\text{b}}+\text{C}'$
$=\frac{1}{\text{b}}\log\big|\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}\big|+\text{C}'$

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