How do you find $int (5x^2+3x-2)/(x^3+2x^2) dx$ using partial fractions?

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How do you find $int (5x^2+3x-2)/(x^3+2x^2) dx$ using partial fractions?

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Answer 1 · 2016-12-11T13:49:40

$2 ln x + 3 ln (x+2) +1/x+C$

Explanation:

Assume that you can write the fraction as $(Ax+B)/x^2+C/(x+2)$ .
That is, $5x^2+3x-2-=(Ax+B)(x+2)+Cx^2$

Then, equating coefficents (or however you prefer)
you get $5=A+C$ , $3=2A+B$ and $-2=2B$ . Hence $B=-1$ , $A=2$ and $C=3$
So now you have $int(2x-1)/x^2+3/x+2 dx$
= $int2/x-1/x^2+3/(x+2)dx$ giving the answer above.

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How do you find $int (5x^2+3x-2)/(x^3+2x^2) dx$ using partial fractions?

1 Answer

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How do you find int (5x^2+3x-2)/(x^3+2x^2) dxint (5x^2+3x-2)/(x^3+2x^2) dx using partial fractions?

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Explanation:

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How do you find $int (5x^2+3x-2)/(x^3+2x^2) dx$ using partial fractions?