Partial fractions to solve integrals

Question

$$\int_0^x\frac{1}{(a-x)(b-x)}\,dx=\frac{1}{b-a}\left( \ln\frac{1}{a-x}-\ln\frac{1}{b-x} \right)$$

I'm trying to figure out how the above fractions are equal to each other. I know they use partial fraction "rules", however I don't quite understand the method in this case. Also we are not looking for the constants $a$ and $b$.

Answer 1

hint

$$\frac {1}{a-x}-\frac {1}{b-x}=$$

$$\frac {b-x-a+x}{(a-x)(b-x)}=$$

$$\frac {b-a}{( a-x)( b-x )} $$

Answer 2

it is $$\frac{1}{(a-x)(b-x)}={\frac {1}{ \left( a-b \right) \left( -a+x \right) }}-{\frac {1}{ \left( a-b \right) \left( -b+x \right) }} $$