How do you integrate $(5x^2+7x-4)/(x^3+4x^2)$ using partial fractions?

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How do you integrate $(5x^2+7x-4)/(x^3+4x^2)$ using partial fractions?

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Answer 1 · 2016-11-04T07:57:44

The answer is $=1/x+2lnx+3ln(x+4)+C$

Explanation:

let's simplify the denominator $x^3+4x^2=x^2(x+4)$
So $(x^2+7x-4)/(x^2(x+4))=A/x^2+B/x+C/(x+4)$

$:.5x^2+7x-4=A(x+4)+Bx(x+4)+Cx^2$
$-4=4A$ $=>$ $A=-1$
coefficients of x, $7=A+4B$ $=>$ $B=2$
Coefficients of x^2, $5=B+C$ $=>$ $C=3$
$:. int((x^2+7x-4)dx)/(x^2(x+4))=int-dx/x^2+int(2dx)/x+int(3dx)/(x+4)$
$=1/x+2lnx+3ln(x+4)+C$

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How do you integrate $(5x^2+7x-4)/(x^3+4x^2)$ using partial fractions?

1 Answer

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How do you integrate (5x^2+7x-4)/(x^3+4x^2)(5x^2+7x-4)/(x^3+4x^2) using partial fractions?

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How do you integrate $(5x^2+7x-4)/(x^3+4x^2)$ using partial fractions?