Hot Integral!

Let us take $\displaystyle f^{ '}(x) = \frac{x}{\sqrt{1+x^2}}$ and $g(x) = ln(x+\sqrt{1+x^2})$ . Then,

$\displaystyle f(x)= \int\frac{x}{\sqrt{1+x^2}} = \frac{1}{2} \int\frac{2x}{\sqrt{1+x^2}} = \frac{1}{2}\int\frac{1}{\sqrt{t}} = \sqrt{t} = \sqrt{1+x^2}$

(Where we have used the substitution $1+x^2 = t$ )

Also,

$\displaystyle g^{ '}(x) = \frac{1}{x+\sqrt{1+x^2}}\cdot \left(1 + \frac{x}{\sqrt{1+x^2}}\right) \Rightarrow g^{ '}(x) = \frac{1}{\sqrt{1+x^2}}$

And, by Integration By Parts, the required integral becomes:

$\displaystyle I \quad =\quad g(x)f(x) -\int { g^{ \prime }\left( x \right) f(x)dx } \\ \displaystyle I\quad =\quad ln(x+\sqrt { 1+{ x }^{ 2 } } )\cdot (\sqrt { 1+{ x }^{ 2 } } )-\int { \frac { 1 }{ { \sqrt { 1+{ x }^{ 2 } } } } \cdot } \sqrt { 1+{ x }^{ 2 } } dx\\ \displaystyle I\quad =\quad ln(x+\sqrt { 1+{ x }^{ 2 } } )\cdot (\sqrt { 1+{ x }^{ 2 } } ) - x + c$

And hence the answer.

Cheers!

Put $x=\sinh(u)$. The integral simplifies greatly, and can then be solved by integration by parts.

$\begin{matrix} let\quad x=\tan { \left( \theta \right) } \Rightarrow dx=\sec ^{ 2 }{ \left( \theta \right) } d\theta & & \color{#D61F06} \text{Trigonometric Substitution} \end{matrix}\\ I=\int { x\cdot \frac { \ln { ( } x+\sqrt { 1+x^{ 2 } } ) }{ \sqrt { 1+x^{ 2 } } } dx } =\int { \tan { \left( \theta \right) } \cdot \frac { \ln { ( } \tan { \left( \theta \right) } +\sqrt { 1+\tan ^{ 2 }{ \left( \theta \right) } } ) }{ \sqrt { 1+\tan ^{ 2 }{ \left( \theta \right) } } } \sec ^{ 2 }{ \left( \theta \right) } d\theta } \\ \begin{matrix} I=\int { \tan { \left( \theta \right) } \cdot \sec { \left( \theta \right) } \cdot \ln { \left( \tan { \left( \theta \right) } +\sec { \left( \theta \right) } \right) } d\theta } & & \color{#3D99F6} \begin{matrix} u=\ln { \left( \tan { \left( \theta \right) } +\sec { \left( \theta \right) } \right) } d\theta & dv=\tan { \left( \theta \right) } \cdot \sec { \left( \theta \right) } \\ du=\frac { \sec { \left( \theta \right) } \left( \tan { \left( \theta \right) } +\sec { \left( \theta \right) } \right) }{ \left( \tan { \left( \theta \right) } +\sec { \left( \theta \right) } \right) } & v= \sec { \left( \theta \right) } \end{matrix} &\color{#D61F06} \text{ integration by parts} \end{matrix}\\ I=\sec { \left( \theta \right) } \cdot \ln { \left( \tan { \left( \theta \right) } +\sec { \left( \theta \right) } \right) } -\int { \sec ^{ 2 }{ \left( \theta \right) } d\theta } \\ I=\sec { \left( \theta \right) } \cdot \ln { \left( \tan { \left( \theta \right) } +\sec { \left( \theta \right) } \right) } -\tan { \left( \theta \right) } +c\\ \begin{matrix} I=\sqrt { 1+x^{ 2 } } \ln { ( } x+\sqrt { 1+x^{ 2 } } )-x+C & & \color{#3D99F6} \begin{matrix} \tan { \left( \theta \right) } =x \\ \sec { \left( \theta \right) } =\sqrt { 1+\tan ^{ 2 }{ \left( \theta \right) } } \end{matrix} \end{matrix}$