How to integrate $\frac{dx}{(x-p)\sqrt {(x-p)(x-q)}} $?

Question

How to integrate $\frac{dx}{(x-p)\sqrt {(x-p)(x-q)}} $ ?

I tried substituting $x=1/t$ but that's making it more complicated. Any suggestions?

Answer 1

An excellent approach would be to substitute $(x-p)=1/t $.Most of the terms can be cancelled out after that.Next using the integration formula for $x^n$ i.e $(x^{n+1})/(n+1) $ is sufficient to reach final answer !

Answer 2

This question is all about substitution. By using $(x-p)(x-q)=(x-\frac{p+q}{2})^2-(\frac{p-q}{2})^2$. Now substitute $x-\frac{p+q}{2}=\frac{p-q}{2}sec(t)$. So $x-p=\frac{p-q}{2}(sec(t)-1)$. Now integral will become $$I=\int \frac{\frac{(p-q)}{2}sec(t)tan(t)dt}{\frac{p-q}{2}(sec(t)-1)|\frac{p-q}{2}|tan(t)}=\int \frac{2dt}{|p-q|(1-cos(t))}$$ So $I$ will be equal to

$$I=-\frac{2cot(\frac{t}{2})}{|p-q|}$$ Put the value of $t$ and get integral in terms of $x$

Answer 3

Hint. By the change of variable $$ u=\sqrt{\frac{x-q}{x-p}},\qquad du=-\frac{(p-q)}2\cdot \frac{dx}{(x-p)\sqrt {(x-p)(x-q)}} $$ one deduces

$$ \int\frac{dx}{(x-p)\sqrt {(x-p)(x-q)}} =-\frac2{(p-q)} \int du=-\frac2{(p-q)} \sqrt{\frac{x-q}{x-p}}+C. $$