Evaluate: $\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{\class{placeholder}{}}}_{\cssId{lower-bound-mathjax}{\class{placeholder}{}}} \ln\left(x+\sqrt{1+x^2}\right)\,\cssId{int-var-mathjax}{\mathrm{d}x}$
As of now, I am able to simplify it to
$\displaystyle\int\ln\left(x\right) + \int\ln\left(\sqrt{x^2+\dfrac{1}{x^2}}+1\right)$
Edit: The answer is $x\ln\left(\sqrt{1+{x^2}}+x\right) - (\sqrt{1+{x^2}}) + C$
It is worth noting that $$\log \left(x + \sqrt{1+x^2}\right) = \sinh^{-1} x.$$ This fact is left as an algebraic exercise for the reader.
Hence, we let $y = \sinh^{-1} x$, so that $x = \sinh y$; then $dx = \cosh y \, dy$ and we have $$\int \sinh^{-1} x \, dx = \int y \cosh y \, dy.$$ Now integration by parts with the choice $$u = y, \quad du = dy, \\ dv = \cosh y \, dv, \quad v = \sinh y,$$ yields $$\int \sinh^{-1} x \, dx = y \sinh y - \int \sinh y \, dy = y \sinh y - \cosh y + C.$$ Then because $$\cosh \sinh^{-1} x = \sqrt{1 + x^2},$$ another algebraic fact left as an exercise for the reader, substituting back gives $$\int \log\left(x + \sqrt{1 + x^2} \right) \, dx = x \log\left(x + \sqrt{1 + x^2} \right) - \sqrt{1+x^2} + C$$ as claimed.
