$\int \ln(1+\sqrt{x})\,dx$ any ideas? $\int \ln(1+\sqrt{x})dx$, any ideas ? 
I tried the substitution : $u = 1+ \sqrt{x}$ but it doesn't sem to work
EDIT : Let $u = 1+\sqrt{x}$ then : $du = \frac{1}{2\sqrt{x}}dx \Leftrightarrow dx = du \cdot 2(u-1)$ Hence we have : $\int \ln(1+\sqrt{x})dx = \int \ln(u) \cdot 2(u-1) du$ But this form doesn't seem to help ? 
 A: You start with
$$\int \ln(1+\sqrt{x}) \, dx$$
then use the substitution $x=u^2$ so that you have
$$\int 2u\ln(u+1) \, du$$
Then use integration by parts to get
$$u^2\ln(u+1)-\int \frac{u^2}{u+1} \, du$$
Split up the $\frac{u^2}{u+1}$ using partial fractions
$$u^2\ln(u+1)-\int (u-1+\frac{1}{u+1}) \, du$$
This is now a pretty straightforward integral. Continue on to get
$$u^2\ln(u+1)-\frac{1}{2}u^2+u-ln(u+1)$$
$$(u^2-1)\ln(u+1)-\frac{1}{2}u^2+u+C$$
Then substitute $u=\sqrt{x}$:
$$(x-1)\ln(\sqrt{x}+1)-\frac{1}{2}x+\sqrt{x}+C$$
Which should be the answer.
A: We can do the first phase by parts:
$$\int \ln(1+\sqrt x)\ \mathrm dx = x \ln(1+\sqrt x) - \frac 12 \int \frac{\sqrt x}{1 + \sqrt x}\ \mathrm dx$$
Let $u = \sqrt x$; then $x = u^2$ and $\mathrm dx = 2u\ \mathrm du$, so:
$$\int \frac{\sqrt x}{1+\sqrt x}\ \mathrm dx = \int \frac{2u^2}{1 + u}\ \mathrm du = \int 2\left(u - \frac u{1+u}\right)\mathrm du = \int 2\left(u - 1 + \frac 1{u+1}\right)\mathrm du = u^2 - 2u + 2\ln(u+1)$$
So, the final formula becomes
$$\int \ln(1 + \sqrt x)\mathrm dx = x\ln(1+\sqrt x) - \frac x2 + \sqrt x - \ln(1+\sqrt x) = (x-1)\ln(1+\sqrt x) + \sqrt x - \frac x2 + C$$
A: \begin{align}
u & = 1 + \sqrt x \\[10pt]
u-1 & = \sqrt x \\[10pt]
(u-1)^2 & = x \\[10pt]
2(u-1)\,du & = dx \\[10pt]
\int \ln(1+\sqrt x) \, dx & = \int (\ln u) \Big(2(u-1)\, du\Big) \\[10pt]
= \int w \, dv & = wv - \int v\,dw = (\ln u)(u-1)^2 - \int (u-1)^2 \left( 
\frac{du} u \right) \\[10pt]
& = (\ln u)(u-1)^2 - \int\left( u - 2 + \frac 1 u \right) \, du = \cdots\cdots
\end{align}
