How do you integrate $int (-2x-3)/(x^2-x)$ using partial fractions?

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Answer 1 · 2016-11-17T16:59:19

The answer is $=3lnx-5ln(x-1)+C$

Let's do the decomposition in partial fractions

The denominator $=x^2-x=x(x-1)$

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

$=(A(x-1)+Bx)/(x(x-1))$

Therefore,
$-2x-3=A(x-1)+Bx$

let $x=1$ , $=>$ $-5=B$

let $x=0$ , $=>$ $-3=-A$ $=>$ $A=3$

So, $(-2x-3)/(x(x-1))=3/x-5/(x-1)$

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

$=3lnx-5ln(x-1)+C$

as $intdx/x=lnx+c$

How do you integrate int (-2x-3)/(x^2-x) int (-2x-3)/(x^2-x) using partial fractions?