Question

$\int \left[\frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}\right]dx=$

• A. ${\left[\log \left(x+\sqrt{1+x^2}\right)\right]}^2+c$
• B. $x\log \left(x+\sqrt{1+x^2}\right)+c$
• C. $\left(\frac{1}{2}\right)\log \left(x+\sqrt{1+x^2}\right)+c$
• D. $\left(\frac{1}{2}\right){\left[\log \left(x+\sqrt{1+x^2}\right)\right]}^2+c$
• E. $\left(\frac{x}{2}\right)\log \left(x+\sqrt{1+x^2}\right)+c$

Answer 1

The correct option is D

$\left(\frac{1}{2}\right){\left[\log \left(x+\sqrt{1+x^2}\right)\right]}^2+c$

Explanation for the correct option:

Finding the value of given integration:

Consider the given equation as,

$I=\int \left[\frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}\right]dx$

Let us assume that,

$$\begin{gathered} \log \left(x+\sqrt{1+x^2}\right)=t \\ \Rightarrow \frac{dx}{\sqrt{1+x^2}}=dt \end{gathered}$$ [differentiating]

Then,

$$\begin{gathered} I=\int tdt \\ I=\frac{t^2}{2}+c \end{gathered}$$

Substitute the value of the $t$ in the above Equation

$$\begin{gathered} I=\frac{{\left(\log \left(x+\sqrt{1+x^2}\right)\right)}^2}{2}+c \\ I=\left(\frac{1}{2}\right){\left[\log \left(x+\sqrt{1+x^2}\right)\right]}^2+c \end{gathered}$$

Hence, the correct answer is Option (D).