Question

If $f\left(x\right)=\int \frac{x^{2}dx}{(1+x^2)(1+\sqrt{1+x^2})}$ and $f\left(0\right)=0$, then the value of $f\left(1\right)$ is

• A. $\log (1+\sqrt{2})$
• B. $\log (1+\sqrt{2})-\frac{\pi }{4}$
• C. $\log (1+\sqrt{2})+\frac{\pi }{4}$
• D. $Noneofthese$

Answer 1

The correct option is B $\log (1+\sqrt{2})-\frac{\pi }{4}$

$f\left(x\right)=\int \frac{x^{2}dx}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}dx\Rightarrow f\left(x\right)=\int \frac{t^2-1}{t^2\left(1+t\right)}\times \frac{tdt}{\sqrt{t^2-1}}\left[\begin{array}{c}id{x}^2=12\\ 2x-dx=2tdt\\ dx=\frac{t}{\sqrt{t^2-1}}dt\end{array}\right]=\int \frac{\left(t-1\right)\left(t+1\right)}{t^2\left(1+t\right)}\times \frac{tdt}{\sqrt{t^2-1}}=\int \frac{t-1}{t\sqrt{t^2-1}}dt=\int \frac{t}{t\sqrt{t^2-1}}dt-\int \frac{1}{t\sqrt{t^2-1}}dt=\int \frac{1}{\sqrt{t^2-1}}dt-\int \frac{1}{t\sqrt{t^2-1}}dt=\ln \left|t+\sqrt{t^2-1}\right|-{\sec }^{-1}t+c$

$f\left(x\right)=\ln \left|\sqrt{1+x^2}+\sqrt{1+x^2-1}\right|-{\sec }^{-1}t+c=In\left|\sqrt{1+x^2}+x\right|-{\sec }^{-1}t+c$

$Atx=0f\left(x\right)=0c=0x=1f\left(x\right)=?$

Therefore,

$f\left(x\right)=In\left|\sqrt{2}+1\right|-{\sec }^{-1}\sqrt{2}=In\left|\sqrt{2}+1\right|-\frac{\pi }{4}$

Hence, this is the answer.