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

Question

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

✓

Answer

We have
$\text{I}=\int\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
Putting $\text{x}=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=\int\frac{-\frac{1}{\text{t}^2}\text{dt}}{\big(1+\frac{1}{\text{t}^2}\big)\sqrt{1-\frac{1}{\text{t}^2}}}$
$=\int\frac{-\frac{1}{\text{t}^2}\text{ dt}}{\frac{(\text{t}^2+1)}{\text{t}^2}\frac{\sqrt{\text{t}^2-1}}{\text{t}}}$
$=-\int\frac{\text{t dt}}{(\text{t}^2+1)\sqrt{\text{t}^2-1}}$
Again Putting $\text{t}^2-1=\text{u}^2$
$2\text{tdt}=2\text{udu}$
$\text{tdt}=\text{udu}$
$\therefore\ \text{I}=-\int\frac{\text{u du}}{(\text{u}^2+2)\text{u}}$
$=-\int\frac{\text{du}}{\text{u}^2+(\sqrt{2})^2}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{u}}{\sqrt{2}}\Big)+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\bigg(\frac{\sqrt{\text{t}^2-1}}{\sqrt{2}}\bigg)+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Bigg(\sqrt{\frac{\frac{1}{\text{x}^2}-1}{2}}\Bigg)+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\bigg(\sqrt{\frac{1-\text{x}^2}{2\text{x}^2}}\bigg)+\text{C}$

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