How do you find $int (11-2x)/(x^2 + x - 2) dx$ using partial fractions?

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Answer 1 · 2016-10-22T05:49:51

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

First factorise the denominator
$x^2+x-2=(x-1)(x+2)$

Then we look for the partial fraction

$(11-2x)/((x-1)(x+2))=A/(x-1)+B/(x+2)$

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

So $11-2x=A(x+2) +B(x-1)$

If $x=1$ $=>$ $11-2=3A+0$
so $A=9/3=3$

If $x=-2$ $=>$ $11+4=0-3B$
so $B=15/-3=-5$

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

Then we can integrate

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

$=3ln(x-1)-5ln(x+2)+C$

Answer 2 · 2016-10-23T02:40:23

Oct 23, 2016

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How do you find int (11-2x)/(x^2 + x - 2) dxint (11-2x)/(x^2 + x - 2) dx using partial fractions?