Question

Evaluate: $\int \frac{dx}{x\sqrt{x^2+c}}$

• A. $\frac{1}{\sqrt{c}}\ln \left|\frac{-\sqrt{c}-\sqrt{c+x^2}}{x}\right|+C$
• B. $\frac{1}{\sqrt{c}}\ln \left|\frac{-\sqrt{c}+\sqrt{c-x^2}}{x}\right|+C$
• C. $\frac{1}{\sqrt{c}}\ln \left|\frac{-\sqrt{c}+\sqrt{c+x^2}}{x}\right|+C$
• D. None of these

Answer 1

The correct option is C $\frac{1}{\sqrt{c}}\ln \left|\frac{-\sqrt{c}+\sqrt{c+x^2}}{x}\right|+C$

Given : $\int \frac{dx}{x\sqrt{x^2+c}}=\int \frac{dx}{x^2\sqrt{1+\frac{c}{x^2}}}$

Let : $\frac{\sqrt{c}}{x}=t$

$-\frac{dt}{\sqrt{c}}=+\left(\frac{dx}{x^2}\right)-\int \frac{dt}{\sqrt{c\sqrt{1+t^2}}}=\frac{1}{\sqrt{c}}\ln |t+\sqrt{t^2+1}|=\frac{1}{\sqrt{c}}\ln |-t+\sqrt{t^2+1}|=\frac{1}{\sqrt{c}}\ln |-\frac{\sqrt{c}}{x}+\sqrt{\frac{c}{x^2}+1}|=\frac{1}{\sqrt{c}}\ln \left|\frac{-\sqrt{c}+\sqrt{c+x^2}}{x}\right|+C$

Hence the correct answer is $\frac{1}{\sqrt{c}}\ln \left|\frac{-\sqrt{c}+\sqrt{c+x^2}}{x}\right|+C$