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

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Answer 1 · 2017-01-19T13:37:33

The answer is $=-1/x+2ln(∣x∣)-ln(∣x+3∣)+C$

Let's perform the decomposition into partial fractions

$(x^2+7x+3)/(x^2(x+3))=A/x^2+B/x+C/(x+3)$

$=(A(x+3)+Bx(x+3)+Cx^2)/(x^2(x+3))$

Equalising the denominators

$x^2+7x+3=A(x+3)+Bx(x+3)+Cx^2$

Let $x=0$ , $=>$ , $3=3A$ , $=>$ , $A=1$

Let $x=-3$ , $=>$ , $-9=9C$ , $=>$ , $C=-1$

Coefficients of $x^2$

$1=B+C$ , $=>$ , $B=1-C=2$

Therefore,

$(x^2+7x+3)/(x^2(x+3))=1/x^2+2/x-1/(x+3)$

So,

$int((x^2+7x+3)dx)/(x^2(x+3))=intdx/x^2+2intdx/x-intdx/(x+3)$

$=-1/x+2ln(∣x∣)-ln(∣x+3∣)+C$

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