Question

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

• A. $I=\log |x-\sqrt{1+x^2}|+{\sin }^{-1}x+c$
• B. $I=\log |x+\sqrt{1+x^2}|+{\sin }^{-1}x+c$
• C. $I=\log |x+\sqrt{1+x^2}|-{\sin }^{-1}x+c$
• D. None of these

Answer 1

The correct option is B $I=\log |x+\sqrt{1+x^2}|+{\sin }^{-1}x+c$

Now,

$\int \frac{\sqrt{1-x^2}+\sqrt{1+x^2}}{\sqrt{1-x^4}}dx$

$=\int \frac{\sqrt{1-x^2}+\sqrt{1+x^2}}{\sqrt{(1-x^2)(1+x^2)}}dx$

$=\int \frac{\sqrt{1-x^2}}{\sqrt{(1-x^2)(1+x^2)}}dx+$$\int \frac{\sqrt{1+x^2}}{\sqrt{(1-x^2)(1+x^2)}}dx$

$=\int \frac{dx}{\sqrt{(1+x^2)}}dx+$$\int \frac{dx}{\sqrt{(1-x^2)}}$

$=\log |x+\sqrt{1+x^2}|+{\sin }^{-1}x+c$ [ Where $c$ is integrating constant]