Question

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

• A. $\ln |x+\sqrt{x^2+1}|+\ln |x-\sqrt{x^2-1}|+C,|x|>1$
• B. $\ln |x+\sqrt{x^2+1}|-\ln |x+\sqrt{x^2-1}|+C,|x|>1$
• C. $2\ln |x+\sqrt{x^2+1}|+\ln |x+\sqrt{x^2-1}|+C,|x|>1$
• D. $2\ln |x+\sqrt{x^2+1}|-\ln |x+\sqrt{x^2-1}|+C,|x|>1$

Answer 1

Answer: (B)

$\ln |x+\sqrt{x^2+1}|-\ln |x+\sqrt{x^2-1}|+C,|x|>1$

Solution:

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

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

where, $x^2-1>0i.e.|x|>1$

$I=ln|x+\sqrt{x^2+1}|-ln|x+\sqrt{x^2-1}|+C,|x|>1$

Here, we know that

$\int \frac{1}{\sqrt{x^2+1}}=ln|x+\sqrt{x^2+1}|+C$ and

$\int \frac{1}{\sqrt{x^2-1}}=ln|x+\sqrt{x^2-1}|+C$

Hence, B is the correct option.