How to integrate it $\int_{0}^{b}\ln(x+\sqrt{x^2+1})dx=?$ I would appreciate if somebody could help me with the following problem: 
Q:How to integrate it 
$$\int_{0}^{b}  \ln(x+\sqrt{x^2+1})dx=?(b>0)$$
 A: Making the change of variables $x=\sinh t$ we get
$$\ln(x+\sqrt{x^2+1})=\ln(\sinh t+\cosh t)=\ln e^{t}=t,$$
and also $dx=\cosh t\, dt$. Can you take it from here?
If not, here is a spoiler:

 Integrate by parts:
 \begin{align} \int t\,d(\sinh t)=t\sinh t-\int \sinh t\,dt=t\sinh t-\cosh t.\end{align}
 Coming back to the initial variable $x$, we get
 \begin{align} x\ln(x+\sqrt{x^2+1})-\sqrt{x^2+1}.\end{align}

An alternative way would be to integrate once by parts to kill the logarithm and then notice that:
$$x\left[\ln(x+\sqrt{x^2+1})\right]'=\frac{x}{x+\sqrt{x^2+1}}\left(1+\frac{x}{\sqrt{x^2+1}}\right)=\frac{x}{\sqrt{x^2+1}}=\left[\sqrt{x^2+1}\right]'.$$
A: There are a lot of ways to get started:


*

*The integrand is a logarithm: maybe the usual trick for using integration by parts will work for this integrand just as well as $\ln x \, dx$?

*We know how to integrate logarithms, but the argument is complicated. We can substitute $u = x + \sqrt{x^2 + 1}$


or even


*

*Maybe we should simplify that radical, and substitute $x = \tan \theta$ or $x = \sinh u$, or a suitable rational substitution if you've seen that trick.


The trick is how to know which to use?
Actually, that's a total lie: the trick is to actually try working through all of the leads you have on a problem to see if it gets you anywhere useful. One of the most important things to understand about problem solving is that, most of the time, you don't actually know if something is going to work before you start trying it. You just have to get out there and try things until you find something that works, or at least converts the problem to another one that you think you can figure out how to solve.
One of the three methods I've mentioned leads to an integrand of a form you should already know how to do.
One of these methods is in the other answer, which also provides an additional observation that may have been nonobvious about how to simplify the result.
I don't think the remaining method leads to an easy solution... but I didn't try very much, and maybe it is a straightforward too.

A fourth method depends on special knowledge: the integrand looks awfully like how the inverse hyperbolic trigonometric functions look when expressed as logarithms (the same with the circular ones, except those involve $\mathbf{i}$). So you could look up which one it is and consult your integral tables -- or use the appropriate method for deriving it on your own, which is the same integration by parts trick as with $\ln x \, dx$ (you don't necessarily have to figure out which inverse function it is to use the IBP trick).
A: The Maple command
with(Student[Calculus1]): IntTutor(ln(x+sqrt(x^2+1)), x); 
does the job by integrating by parts. It outputs that.
