How do you use partial fractions to find the integral $int (2x-3)/(x-1)^2dx$ ?

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Answer 1 · 2016-12-09T09:58:34

$int frac (2x-3) ((x-1)^2) dx = 2 ln|x-1| +1/(x-1)$

As the denominator of the integrand is already factorized we can quickly decompose it using partial fractions:

$frac (2x-3) ((x-1)^2) = A/(x-1) + B/((x-1)^2)$

$frac (2x-3) ((x-1)^2) = frac (A(x-1) + B) ((x-1)^2) = frac (Ax-A + B) ((x-1)^2)$

Hence:

$A=2$
$B=-1$

and:

$int frac (2x-3) ((x-1)^2) dx = 2 int dx/(x-1) - int dx/((x-1)^2)$

$int frac (2x-3) ((x-1)^2) dx = 2 ln|x-1| +1/(x-1)$

How do you use partial fractions to find the integral int (2x-3)/(x-1)^2dxint (2x-3)/(x-1)^2dx?