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?
 A: Here is a solution not working with hyperbolic functions. Now updated with some more details.
Integrating by parts, and using that (by the chain rule, and writing the expression inside square brackets on common denominator)
$$
\begin{aligned}
D\log(\sqrt{x}+\sqrt{x+1})&=\frac{1}{\sqrt{x}+\sqrt{x+1}}D\bigl(\sqrt{x}+\sqrt{x+1}\bigr)\\
&= \frac{1}{\sqrt{x}+\sqrt{x+1}}\Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\Bigr]\\
&=\frac{1}{\sqrt{x}+\sqrt{x+1}}\frac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x}\sqrt{x+1}}\\
&=\frac{1}{2\sqrt{x}\sqrt{x+1}},
\end{aligned}
$$
we find that
$$
\int \log(\sqrt{x}+\sqrt{x+1})\,dx=x\log(\sqrt{x}+\sqrt{x+1})-\int \frac{\sqrt{x}}{2\sqrt{x+1}}\,dx
$$
Then, integrating by parts again, rewriting, splitting, and using the derivative on top,
$$
\begin{aligned}
\int \frac{\sqrt{x}}{2\sqrt{x+1}}\,dx&=\sqrt{x}\sqrt{x+1}-\int \frac{\sqrt{x+1}}{2\sqrt{x}}\,dx\\
&=\sqrt{x}\sqrt{x+1}-\int\frac{x+1}{2\sqrt{x}\sqrt{x+1}}\,dx\\
&=\sqrt{x}\sqrt{x+1}-\int\frac{\sqrt{x}}{2\sqrt{x+1}}\,dx-\log(\sqrt{x}+\sqrt{x+1}).
\end{aligned}
$$
Moving the integral to the left-hand side, we find that
$$
\int \frac{\sqrt{x}}{2\sqrt{x+1}}\,dx=\frac{1}{2}\sqrt{x}\sqrt{x+1}-\frac{1}{2}\log(\sqrt{x}+\sqrt{x+1}),
$$
and so, finally, one primitive is given by
$$
\int\log(\sqrt{x}+\sqrt{x+1})\,dx = \Bigl(x+\frac{1}{2}\Bigr)\log(\sqrt{x}+\sqrt{x+1})-\frac{1}{2}\sqrt{x}\sqrt{x+1}.
$$
A: 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.
A: In fact, $u\ge 1$ in your integral (due to $\sqrt{u-1}$ present). So the substitution $u = \cos^2 t$ is not possible. However, you can use the hyperbolic substitution $u = \cosh t$ (of $x = \sinh t$ immediately) to get the same result as Walpha.
A: Recall that $$\sinh t =\frac{e^t-e^{-t}}{2}\text{, and } \cosh t =\frac{e^t+e^{-t}}{2} $$
Now note that \begin{align}
\cosh^2 t -\sinh^2 t&=\Big(\frac{e^t+e^{-t}}{2}\Big)^2-\Big(\frac{e^t+e^{-t}}{2}\Big)^2\\
&=\frac{e^{2t}+e^{-2t}+2}{2^2}-\frac{e^{2t}+e^{-2t}-2}{2^2}\\
&=\frac{2+2}{2^2}=1
\end{align}
Hence if in your integral you choose $x=\sinh ^2 t$ you get \begin{align}\sqrt x + \sqrt{x+1}&=\sinh t + \sqrt{\sinh^2t+1}\\
&=\sinh t + \sqrt{\cosh^2t}\\
&=\sinh t + \cosh t\\
&=\frac{e^t-e^{-t}}{2}+\frac{e^t+e^{-t}}{2}\\
&=e^t
\end{align}
.. this, hopefully, clarifies things for you.
