calculus - How can I calculate the following integral (partial fractions)?

$\int_{\frac{3}{2}}^{2} \sqrt\frac{x-1}{3-x}\,dx$

I tried to integrate at first with $u= \sqrt \frac{x-3}{x-1}$ and after it I get a very complicated integration by partial fractions, can I solve it with an other way too? I think it has to be a much easier method, I appreciate any help.

Answer

You idea was very good.

Considering

$$I=\int \sqrt\frac{x-1}{3-x}\,dx$$ $$u= \sqrt\frac{x-1}{3-x}\implies x=\frac{3 u^2+1}{u^2+1}\implies dx=\frac{4 u}{\left(u^2+1\right)^2}du$$ So $$I=\int\frac{4 u^2}{\left(u^2+1\right)^2}dt=4\int\frac{du}{u^2+1}-4\int\frac{du}{\left(u^2+1\right)^2}$$ The first one is simple; for the second one, use integration by parts.

I suppose that you did not work enough the simplification of $dx$.