Question

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

• A. $cos{h}^{-1}x+{sin}^{-1}x+c$
• B. $cos{h}^{-1}x+co{s}^{-1}x+c$
• C. $sin{h}^{-1}x+si{n}^{-1}x+c$
• D. $sin{h}^{-1}x+co{s}^{-1}x+c$

Answer 1

The correct option is C $sin{h}^{-1}x+si{n}^{-1}x+c$

Given,

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

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

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

$=\sin h^{-1}x+{\sin }^{-1}x+c$