Mathematics

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


ANSWER

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


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
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