Evaluate the following intregals: x-1 x^2+1 dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\text{x}-1}{\sqrt{\text{x}^2+1}}\text{dx}$

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Answer

$\int\frac{\text{x}-1}{\sqrt{\text{x}^2+1}}\text{dx}$
$=\int\frac{\text{x}}{\sqrt{\text{x}^2+1}}-\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}=\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2+1}}\text{dx}-\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}$
$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}-\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}=\frac{1}{2}(2\sqrt{\text{t}})-\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}\\=\sqrt{\text{t}}-\text{In}\big|\text{x}+\sqrt{\text{x}^2+1}\big|+\text{c}$
$=\sqrt{\text{x}^2+1}-\text{In}\big|\text{x}+\sqrt{\text{x}^2+1}\big|+\text{c}$

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