How to compute $\int{\ln(\sqrt{x}+\sqrt{x+1})dx}$? The integral I am trying to work out is
$$\int{\ln(\sqrt{x}+\sqrt{x+1})dx}$$
So first step is substitute $u=x+1$:
$$\int{\ln(\sqrt{u-1}+\sqrt{u})du}.$$
Now substitute $u=\cos^2(t)$.
$\int{-\ln(\sqrt{-\sin^2(t)}+\sqrt{\cos^2(t)})2\sin(t)\cos(t)dt}$. This works out to $\int{-\ln(\cos(t)+i\sin(t))\sin(2t)dt}$. Remarkably, we obtain $\int{-\ln(e^{it})\sin(2t)dt}$ using Euler's identity.
Now we have $\int{-it\sin(2t)dt}$ and filling in reverse substitution $t=\arccos(\sqrt{x+1})$ after performing partial integration eventually yields me $0.5i\arccos(\sqrt{x+1})(2x+1)+0.5\sqrt{x^2+x}$.
I tried it on wolfram alpha and it gave a different result with hyperbolic functions. So that was not too useful to check this answer with. I feel this is not the right answer due to the imaginary inverse cosine factor. Where did it go wrong?

HINT...The integral is equivalent to $$I=\int\operatorname{arsinh} \sqrt{x}dx$$
Therefore it would make sense to deal with hyperbolic functions here. 
You could start with the substitution $$\sqrt{x}=\sinh u$$
This would give $$I=\int u\sinh 2udu$$
This can be done by parts.
