Integrate $\frac{x\ln(x+\sqrt{x^{2}+1})}{\sqrt{x^{2}+1}}$

Question

$$\int \frac{x\ln\left(x+\sqrt{x^{2}+1}\right)}{\sqrt{x^{2}+1}}dx$$

I honestly have been struggling with this integral for 2 hours straight and none of my methods (including substitution and integration by parts) have worked. I desperately need help...

Answer 1

Set $u=x+\sqrt{x^2+1}$ so $$x^2+1=(u-x)^2=u^2-2ux+x^2\implies x=\frac{u^2-1}{2u}$$and $$\frac{du}{dx}=1+\frac{x}{\sqrt{x^2+1}}=\frac{u}{u-x}=\frac{2u^2}{u^2+1}.$$Since $\frac{x}{\sqrt{x^2+1}}=\frac{x}{u-x}=\frac{u^2-1}{u^2+1}$, the integral is $$\int\frac{u^2-1}{2u^2}\ln u du=\frac{1}{2}(\int \ln u du-\int \ln y dy)$$with $y=1/u$, so the result is $$\frac{1}{2}((u+\frac{1}{u})\ln u-u + \frac{1}{u})+C.$$Using $\frac{1}{u}=\sqrt{x^2+1}-x$, we can simplify this to $$\sqrt{x^2+1}\ln (x+\sqrt{x^2+1})-x+C.$$

Answer 2

$$ \int \frac{x \ln(x + \sqrt{x^{2} + 1})}{ \sqrt{x^{2} + 1} } dx $$

let $u = x+\sqrt{x^{2} + 1}$, then $du = 1+\frac{x}{ \sqrt{x^{2}+1}} dx \implies dx = \frac{\sqrt{x^{2} + 1}}{x + \sqrt{x^{2} + 1}} du$

$$ \int \frac{x \ln(x + \sqrt{x^{2} + 1})}{ \sqrt{x^{2} + 1} } dx = x \int \frac{ \ln(u) }{u} du - \int \left( \int \frac{ \ln(u) }{u} du \right) dx$$

Can you proceed..?

i the end you have to calculate this though ...

$$ \int \ln^{2}(x + \sqrt{x^{2} + 1}) dx $$

Answer 3

Noting $d\sqrt{x^{2}+1}=\frac{x}{\sqrt{x^2+1}}dx$ and then by integration by parts, one has \begin{eqnarray} &&\int \frac{x\ln\left(x+\sqrt{x^{2}+1}\right)}{\sqrt{x^{2}+1}}dx\\ &=&\int \ln\left(x+\sqrt{x^{2}+1}\right)d\sqrt{x^{2}+1}\\ &=&\sqrt{x^{2}+1}\ln\left(x+\sqrt{x^{2}+1}\right)-\int\sqrt{x^2+1}d\ln\left(x+\sqrt{x^{2}+1}\right)\\ &=&\sqrt{x^{2}+1}\ln\left(x+\sqrt{x^{2}+1}\right)-\int\sqrt{x^2+1}\frac{1}{x+\sqrt{x^{2}+1}}\left(1+\frac{x}{\sqrt{x^2+1}}\right)dx\\ &=&\sqrt{x^{2}+1}\ln\left(x+\sqrt{x^{2}+1}\right)-\int dx\\ &=&\sqrt{x^{2}+1}\ln\left(x+\sqrt{x^{2}+1}\right)-x+C\\ \end{eqnarray}