How do you evaluate the integral $int (2x-3)/((x-1)(x+4))$ ?

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Answer 1 · 2018-03-08T22:48:46

$int (2x-3)/((x-1)(x+4)) dx = -1/5 ln abs (x-1)+11/5 ln abs (x+4) +C$

Use partial fractions decomposition:

$(2x-3)/((x-1)(x+4)) = A/(x-1)+B/(x+4)$

$2x-3 = A(x+4)+B(x-1)$

$2x-3 = (A+B)x +4A-B$

${(A+B=2),(4A-B=-3):}$

${(5A=-1),(B=-A+2):}$

${(A=-1/5),(B=11/5):}$

$(2x-3)/((x-1)(x+4)) = -1/(5(x-1))+11/(5(x+4))$

$int (2x-3)/((x-1)(x+4)) dx = -1/5 int dx/(x-1)+11/5 int dx/(x+4)$

$int (2x-3)/((x-1)(x+4)) dx = -1/5 ln abs (x-1)+11/5 ln abs (x+4) +C$

How do you evaluate the integral int (2x-3)/((x-1)(x+4))int (2x-3)/((x-1)(x+4))?