Find $$\int\frac{1}{x\sqrt{x-1}}\ dx.$$
I have attempted to use the $A/x + B/\sqrt{x-1}$ method. That does not work. I have tried a substitution with $t= \sqrt{x-1}$ and $t^2 +1 = x$. Finding the $dt$ and exchanging it with $dx$ creates another square root in the denominator.
On
The choice you may have been looking for is a "rationalizing substitution", $t^2 = x - 1$, which has the differential $2t \, dt = dx$. The integral then becomes
$$ \int \frac{1}{x\sqrt{x-1}}\, dx \ \ \rightarrow \ \ \int \frac{2 t \, dt}{ (t^2 \ + \ 1) \cdot t} = \int \frac{2}{t^2 + 1 } \, dt,$$
which also leads to the arctangent anti-derivative Nehorai shows.
On
Given that the title of the OP is "Integral of $\frac{1}{x\sqrt{x-1}}$ using partial fractions," I thought it might be instructive to present a solution using partial fraction expansion. To that end, we proceed.
Now, other answers have already been posted wherein enforcing the substitution $x=t^2+1$, the result of the integration was expressed as
$$\int\frac{1}{x\sqrt{x-1}}\,dx=2\int \frac{1}{t^2+1}\,dt \tag 1$$
We now proceed under the assumption that one doesn't recognize the right-hand side of $(1)$ as $2\arctan(t)$, and one insists on evaluating the integral using partial fraction expansion. Then, we can write the right-hand side of $(1)% as
$$\begin{align} 2\int \frac{1}{t^2+1}\,dt&=\frac1i \int \left(\frac{1}{t-i}-\frac{1}{t+i}\right)\,dt\\\\ &=i \log(t+i)-i\log(t-i)+C\\\\ &=\frac i2\log(t^2+1)-\arctan(1/t)-\frac i2\log(t^2+1)-\arctan(1/t)+C'\\\\ &=-2 \arctan(1/t)+C'\\\\ &=2\arctan(t)+C'' \end{align}$$
as expected!
Substitute $t=x-1$ and $dt=dx$
$$=\int\frac{dt}{\sqrt t(t+1)}$$
Substitute $\nu=\sqrt t$ and $d\nu=\frac{dt}{2\sqrt t}$
$$=2\int\frac{d\nu}{\nu^2+1}=2\arctan \nu+\mathcal C=2\arctan \sqrt t+\mathcal C=\color{red}{2\arctan (\sqrt{x-1})+\mathcal C}$$