# Partial fractions to solve integrals [closed]

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

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 $b$.

## closed as off-topic by Namaste, Simply Beautiful Art, Did, eranreches, zz20sDec 22 '17 at 21:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Simply Beautiful Art, Did, eranreches, zz20s
If this question can be reworded to fit the rules in the help center, please edit the question.

• You have an $x$ in the boundary and an $x$ in the integrand. Typically this isn't OK. You probably want the bound to be a different variable, like $t$ or something. Moreover, you may want to include the question written out, as other users may downvote you for attaching an image. – Alfred Yerger Dec 9 '17 at 20:30

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