Question

Solve:

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

• A. $\log (x+\sqrt{1+x^2})+{\sin }^{-1}x+c$
• B. ${\sin }^{-1}x-{\sin }^{-1}x+c$
• C. ${\cos }^{-1}-{\sin }^{-1}x+c$
• D. ${\tan }^{-1}x+{\sin }^{-1}x+c$

Answer 1

Answer: (A)

$\log (x+\sqrt{1+x^2})+{\sin }^{-1}x+c$

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

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

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

$=\log (x+\sqrt{1+x^2})+{\sin }^{-1}x+c$