How to compute $\int_{0}^{(e-1)^2}{\ln(\sqrt{x}+1)} \,\mathrm dx $? I have a problem with this integral.
$$\int\limits_{0}^{(e-1)^2}\!\! \left({\ln(\sqrt{x}+1)} \right)\,\mathrm dx $$
I applied the substitution method $t = \sqrt{x}+1$, $2t = dx$
I changed integration interval from $0 \to (e-1)^2$ to $1 \to e$
$$\int\limits_{1}^{e}\!\! \left(2t\,{\ln(t}) \right)\,\mathrm dt $$
Then I worked for integration by parts
$$={t^2 \ln t}-\int\limits \left (t \right)\,\mathrm dt$$
$$={t^2 \ln t-\frac{t^2}{2}+C} $$
completing the exercise I understand the following result $\frac{e^2+1}{2}$ , which it is obviously wrong. 
Can you correct my mistake? 
 A: There was an error in the substitution:
$$t=\sqrt{x}+1 \implies dt =\frac{1}{2\sqrt{x}}dx=\frac{1}{2(\sqrt{x}+1)-2}dx=\frac{1}{2t-2}dx \implies dx=2(t-1)dt$$ So 
$$\int_{1}^e2(t-1)\ln(t)\,dt=\int_{1}^e\left((t-1)^2\right)'\ln(t)\,dt=[(t-1)^2\ln(t)]_{1}^{e}-\int_{1}^e\frac{(t-1)^2}{t}\,dt$$ I assume you can take it from here.
A: First use the substitution $t=\sqrt x$. This gives $dt=\frac{1}{2}x^{-\frac{1}{2}}dx$, or $dx=2\sqrt x\ dt=2t\ dt$.
Re-evaluate the limits next. When $x=0$, $t=0$, and when $x=(e-1)^2$, $t=e-1$.
The new integral is
$$\int_0^{e-1} 2t \ln(t+1)dt$$
Finally we integrate by parts: $u=\ln(t+1)$ and $dv=2tdt$. Then $du=\frac{1}{t+1}dt$ and $v=t^2$.
Now we have
$$t^2\ln(t+1)+\int\frac{t^2}{t+1}dt$$
You can integrate by parts again, etc.
A: In order to solve 
$$\int_0^{(e-1)^2}{(\ln(\sqrt{x}+1))\mathrm{d}x},$$
We will now use the substitution $\mathrm{u}=\sqrt{x}+1$, or $x=(\mathrm{u}-1)^2$, which provides $\mathrm{d}x = 2(\mathrm{u}-1)\mathrm{du}$. 
From this, we obtain
$$\int_0^{(e-1)^2}{(2(\mathrm{u}-1)\ln(\mathrm{u}))\mathrm{du}}.$$
Using integration by parts with $g'(\mathrm{u})=\ln(\mathrm{u})$ and $h(\mathrm{u})=2(\mathrm{u}-1)$, we can calculate $g(\mathrm{u})=\mathrm{u}\ln(\mathrm{u})-\mathrm{u}$ and $h'(\mathrm{u})=2$. Now, we have
$$\int_0^{(e-1)^2}{(2(\mathrm{u}-1)\ln(\mathrm{u}))\mathrm{du}} = \left. 2(\mathrm{u}-1)(\mathrm{u}\ln(\mathrm{u})-\mathrm{u})\right\vert_{0}^{(e-1)^2}-\int_0^{(e-1)^2}{(2(\mathrm{u}\ln(\mathrm{u})-\mathrm{u}))\mathrm{du}}.$$
Wait, something seems familiar. See it? The second term in the right hand side is the same as the integral we're trying to calculate. We'll use this to our advantage by substituting $y$ is equal to that big nasty integral. Our equation is now
$$y=\left. 2(\mathrm{u}-1)(\mathrm{u}\ln(\mathrm{u})-\mathrm{u}) \right\vert_{0}^{(e-1)^2} -y.$$
Adding $y$ to both sides, our equation is now
$$2y=\left. 2(\mathrm{u}-1)(\mathrm{u}\ln(\mathrm{u})-\mathrm{u}) \right\vert_{0}^{(e-1)^2} ,$$
which is the same as
$$y=\left. (\mathrm{u}-1)(\mathrm{u}\ln(\mathrm{u})-\mathrm{u}) \right\vert_{0}^{(e-1)^2}.$$
I think you've got this from here. Just substitute and simplify! ;-)
If I've made some errors, somebody please point them out.
