Question

If $\int \frac{\sqrt{x^2+1}}{1-x^2}dx=\frac{1}{\sqrt{2}}f\left(x\right)-g\left(x\right)+C$, where $C$ is a constant of integration, then

• A. $f\left(x\right)=\ln \left|\frac{\sqrt{1+x^2}+\sqrt{2}x}{\sqrt{1+x^2}-\sqrt{2}x}\right|$
• B. $g\left(x\right)=\ln |x-\sqrt{1+x^2}|$
• C. $f\left(x\right)=\ln \left|\frac{\sqrt{1+x^2}-\sqrt{2}x}{\sqrt{1+x^2}+\sqrt{2}x}\right|$
• D. $g\left(x\right)=\ln |x+\sqrt{1+x^2}|$

Answer 1

Answer: (A), (D)

The correct options are
A $f\left(x\right)=\ln \left|\frac{\sqrt{1+x^2}+\sqrt{2}x}{\sqrt{1+x^2}-\sqrt{2}x}\right|$
D $g\left(x\right)=\ln |x+\sqrt{1+x^2}|$

Let $I=\int \frac{\sqrt{x^2+1}}{1-x^2}dx$
$\Rightarrow I=\int \frac{\sqrt{x^2+1}}{1-x^2}\cdot \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}dx=\int \frac{1+x^2}{(1-x^2)\sqrt{1+x^2}}dx=\int \frac{2-(1-x^2)}{(1-x^2)\sqrt{1+x^2}}dx=2\int \frac{1}{(1-x^2)\sqrt{1+x^2}}dx-\int \frac{1}{\sqrt{1+x^2}}dx=2\int \frac{1}{(1-x^2)\sqrt{1+x^2}}dx-\ln |x+\sqrt{1+x^2}|+C_1$

Now, let $I_1=2\int \frac{1}{(1-x^2)\sqrt{1+x^2}}dx$
Put $x=\tan \theta \Rightarrow dx=(1+{\tan }^2\theta )d\theta$
So, $I_1=2\int \frac{1+{\tan }^2\theta }{(1-{\tan }^2\theta )\sec \theta }d\theta$
$=2\int \frac{\cos \theta }{\cos 2\theta }d\theta =2\int \frac{\cos \theta }{1-2{\sin }^2\theta }d\theta$

Put $\sin \theta =t\Rightarrow dt=\cos \theta d\theta$
So, $I_1=2\int \frac{1}{1-2t^2}dt=\frac{1}{\sqrt{2}}\ln \left|\frac{1+\sqrt{2}t}{1-\sqrt{2}t}\right|$
Now, $x=\tan \theta \Rightarrow \sin \theta =\frac{x}{\sqrt{1+x^2}}=t$
So, $I_1=\frac{1}{\sqrt{2}}\ln \left|\frac{\sqrt{1+x^2}+\sqrt{2}x}{\sqrt{1+x^2}-\sqrt{2}x}\right|$

Hence, $f\left(x\right)=\ln \left|\frac{\sqrt{1+x^2}+\sqrt{2}x}{\sqrt{1+x^2}-\sqrt{2}x}\right|$ and $g\left(x\right)=\ln |x+\sqrt{1+x^2}|$