Derive reduction formula for $\int \ln^n(x+\sqrt{x^2+1}) \, dx $ I wrote something like $$I=\int(x)' \ln^n {\left(x+\sqrt{x^2+1} \right)} \, dx$$ and $$I=x\ln(x+\sqrt{x^2+1})-n \displaystyle \int x \frac{\ln^{n-1} {\left(x+\sqrt{x^2+1} \right)}}{\sqrt{x^2+1}}dx.$$
I noticed that ${\left[\ln{{\left(x+\sqrt{x^2+1} \right)}}\right]}'=\dfrac{1}{1+x^2},$ but I don't know what I should do next.
 A: Recognize  $\ln (x+\sqrt{x^2+1})=\text{arcsinh}x = t$ and rewrite the integral as
$$I_n=\int \ln^n(x+\sqrt{x^2+1}) \, dx
= \int t^n \cosh t dt
$$
Integrate by parts twice to reduce to
$$I_n = t^n \sinh t - n t^{n-1} \cosh t +n(n-1)I_{n-2}
$$
or
$$I_n = x\>\text{arcsinh}^nx-n \sqrt{x^2+1}\text{arcsinh}^{n-1}x+n(n-1)I_{n-2}
$$
A: Let's use integration by parts.
$\int u_1u_2 \,dx=u_1 \int u_2\,dx-\int u_1' (\int u_2\,dx)\,dx$ 
Sometimes choosing $u_1,u_2$ in certain ways makes integration by parts simpler. In this case,  I'll take logarithmic function as $u_1$ and $1$ as $u_2$ to start with.
Let
$\begin{align} I_n=&\int \ln^n(x+\sqrt{x^2+1}) \times 1 \, dx \\=&\ln^n(x+\sqrt{x^2+1})x-n\int\ln^{n-1}(x+\sqrt{x^2+1})\frac{1}{x+\sqrt{x^2+1}}(1+\frac x{\sqrt{x^2+1}})x \, dx\\=&\ln^n(x+\sqrt{x^2+1})x-n\int \ln^{n-1}(x+\sqrt{x^2+1})\frac x{\sqrt{x^2+1}} \, dx\end{align}$ $\tag 1$ 
Second integral is
$
\int \ln^{n-1}(x+\sqrt{x^2+1})\frac x{\sqrt{x^2+1}} \, dx 
=\ln^{n-1}(x+\sqrt{x^2+1})\sqrt{x^2+1}-\\(n-1)\int\ln^{n-2}(x+\sqrt{x^2+1})
\frac 1{x+\sqrt{x^2+1}}(1+\frac x{\sqrt{x^2+1}})\sqrt{x^2+1} \,dx 
\\=\ln^{n-1}(x+\sqrt{x^2+1})\sqrt{x^2+1}-(n-1)I_{n-2}$ 
So $(1)$ becomes: $I_n=\ln^n(x+\sqrt{x^2+1})x-n\ln^{n-1}(x+\sqrt{x^2+1})\sqrt{x^2+1}+n(n-1)I_{n-2}$, where $n\ge 2$
