Question

If $f\left(x\right)=\int \frac{1}{x-\sqrt{x^2+1}}$ and $f\left(0\right)=\frac{1+\sqrt{2}}{2}$, then $f\left(1\right)$ is equal to

• A. $\log \left(\sqrt{\sqrt{2}-1}\right)$
• B. $\frac{-1}{\sqrt{2}}$
• C. $1+\sqrt{2}$
• D. $\frac{1}{2}\log (1+\sqrt{2})$

Answer 1

The correct option is A $\log \left(\sqrt{\sqrt{2}-1}\right)$

By rationalizing the integrand, the given integral can be written as

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

$f\left(x\right)=-\frac{x^2}{2}-\frac{x}{2}\sqrt{x^2+1}-\frac{1}{2}\log |x+\sqrt{x^2+1}|+c$

Given, $f\left(0\right)=\frac{1+\sqrt{2}}{2}+c$

$\therefore c=\frac{1+\sqrt{2}}{2}$

And,

$f\left(1\right)=-\frac{1}{2}-\frac{\sqrt{2}}{2}-\frac{1}{2}\log |1+\sqrt{2}|+(\frac{1}{2}+\frac{1}{\sqrt{2}})$

$f\left(1\right)=-\frac{1}{2}\log |1+\sqrt{2}|=\frac{1}{2}\log (\sqrt{2}-1)$

$f\left(1\right)=\log \left(\sqrt{\sqrt{2}-1}\right)$