Mathematics

# Solve $\displaystyle \int \dfrac{\sqrt{1 - x^2} + \sqrt{1 + x^2}}{\sqrt{1 - x^4}}dx$;

$I=\log|x+\sqrt{1+x^2}|+\sin^{-1}x+c$

##### SOLUTION
Now,
$\displaystyle \int \dfrac{\sqrt{1 - x^2} + \sqrt{1 + x^2}}{\sqrt{1 - x^4}}dx$
$=\displaystyle \int \dfrac{\sqrt{1 - x^2} + \sqrt{1 + x^2}}{\sqrt{(1 - x^2)(1+x^2)}}dx$
$=\displaystyle \int \dfrac{\sqrt{1 - x^2} }{\sqrt{(1 - x^2)(1+x^2)}}dx+$$\displaystyle \int \dfrac{\sqrt{1 + x^2}}{\sqrt{(1 - x^2)(1+x^2)}}dx =\displaystyle \int \dfrac{dx }{\sqrt{(1+x^2)}}dx+$$\displaystyle \int \dfrac{dx}{\sqrt{(1 - x^2)}}$
$=\log|x+\sqrt{1+x^2}|+\sin^{-1}x+c$ [ Where $c$ is integrating constant]

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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Let a function $f:R\rightarrow R$ be defined as $f\left( x \right) =x+\sin { x }$. The value of $\int _{ 0 }^{ 2\pi }{ { f }^{ -1 }(x) } dx$ will
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