$\int \frac{1}{x^2\sqrt{x-1}}\; dx$ I am trying to do the following integral
$$
\int \frac{1}{x^2\sqrt{x-1}}\; dx
$$
My first thought is to do a substitution $u = x - 1$ so that I get
$$
\int \frac{1}{(u + 1)^2\sqrt{u}}\; du
$$
The maybe another substitution $t = \sqrt{u}$, so that I get
$$
\int \frac{1}{(t^2 + 1)^2} 2\;dt
$$
This might then be a trig. sub., but I don't think it is supposed to be this hard. Help please.
 A: You can combine your two substitutions into one by setting $x-1=t^2$.
$$2\int \frac{1}{(t^2+1)^2} dt$$
Now letting $t=\tan \theta$, the above simplifies to
$$2\int \frac{\sec^2 \theta}{\sec^4 \theta} \, d\theta =\int 2\cos^2 \theta \, d\theta =\int 1+\cos 2\theta \, d\theta$$
$$=\theta + \frac{\sin 2\theta}{2}+C$$
It remains to convert the above into an expression of $x$.
A: After your calculations, you just have to do the substitution $t = \tan(s)$ and it is straightforward
A: Hint:
Set $t=\sqrt{x-1}$, so that $x^2=1+t^2$ and therefore $\;\mathrm dx=2t\,\mathrm dt$.
The integral becomes $$\int\frac{2t\,\mathrm dt}{(1+t^2)^2 t}=2\int\frac{\mathrm dt}{(1+t^2)^2}.$$
Now the integrals $\;I_n=\displaystyle\int\frac{\mathrm dt}{(1+t^2)^n} $ are standard and computed recursively in the following way:

*

*$I_1=\arctan t$.

*For the inductive step, use integration by parts with $I_{n-1}$, setting
$$u=\frac1{(1+t^2)^{n-1}}\enspace\mathrm dv=\mathrm dt,\enspace\text{whence }\quad\mathrm du=-\frac{2(n-1)t\,\mathrm dt}{(1+t^2)^n},\quad v=t,$$
and ultimately
\begin{align}
I_{n-1}&=\frac t{(1+t^2)^{n-1}}+2(n-1)\int\frac {t^2\,\mathrm dt}{(1+t^2)^n}=\frac t{(1+t^2)^n}+2(n-1)\int\frac {(1+t^2)-1}{(1+t^2)^n}\,\mathrm dt \\
&=\frac t{(1+t^2)^{n-1}} +2(n-1)(I_{n-1}-I_n) \\
\implies I_n&= \frac t{2(n-1)(1+t^2)^n}+\frac{2n-3}{2n-2}I_{n-1}.
\end{align}
A: A different approach to calculate
$$\int\frac{1}{(1+t^2)^2}dt$$
is the following:
$$\int\frac{1}{(1+t^2)}\cdot\frac{1}{(1+t^2)}dt$$
$$\int\frac{1}{(1+t^2)}\cdot\frac{d}{dt}\arctan(t)dt$$
$$\int\frac{1}{(1+t^2)}\cdot d[\arctan(t)]$$
$$\int \cos^2(\arctan(t))d[\arctan(t)]=\int \cos^2(y)dy$$
