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

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, zz20s Dec 22 '17 at 21:32

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    $\begingroup$ 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. $\endgroup$ – Alfred Yerger Dec 9 '17 at 20:30


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

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

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

  • $\begingroup$ I understand what you do but how can you derive the first expression? I mean the two partial fractions in your comment? $\endgroup$ – AGR Dec 15 '17 at 8:05

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


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