How to determine this integral Let's take:
$$\int \frac{1}{x\sqrt{x+1}} \ dx  $$
I tried solve this by four hour, so I am asking for help
 A: Here is a fairly common way to do this. We see that
\begin{equation}
\int \frac{1}{x\sqrt{x + 1}} dx = \int \frac{1}{(x + 1 - 1)\sqrt{x+1}}dx,
\end{equation}
which we can re-write as
\begin{equation}
\int\frac{1}{\left((\sqrt{x+1})^2 - 1\right)\sqrt{x+1}}dx.
\end{equation}
From here, we can set $u = \sqrt{x+1}$. Then we have
\begin{equation}
du = \frac{1}{2}\frac{1}{\sqrt{x+1}}
\end{equation}
so that the above integral is
\begin{equation}
\int \frac{2}{u^2 - 1}du.
\end{equation}
The above is a known integral which evaluates to
\begin{equation}
\int \frac{2}{u^2 - 1}du = -2 \tanh^{-1}(u) + C,
\end{equation}
whereupon we substitute $u$ out to get
\begin{equation}
-2 \tanh^{-1}(\sqrt{x+1}) + C.
\end{equation}
A: $\int{\frac{1}{x\sqrt{x+1}}}dx$
You can make substitution $u^2 = x +1$
$x = u^2 - 1$
$2u\cdot du = dx$
Then you will get
$\int{\frac{2u}{(u^2-1)u}}du = \int{\frac{2}{(u^2-1)}}du = \int{\frac{2 \cdot(-1)}{(u^2-1)\cdot (-1)}}du = -\int{\frac{2}{(1-u^2)}}du = -2arctanh(u) + c = -2arctanh(\sqrt{x+1}) +c$
