How do you integrate (x^3+25)/(x^2+4x+3) using partial fractions?

2 Answers
Jim G.
Jul 8, 2017

12ln|x+1|+ln|x+3|+c

Explanation:

"factorising the numerator"

(x^3+25)/((x+1)(x+3)

rArr(x^3+25)/((x+1)(x+3))=A/(x+1)+B/(x+3)

"multiply through by " (x+1)(x+3)

rArrx^3+25=A(x+3)+B(x+1)

"using the "color(blue)"cover up method"

x=-3to-2=-2BrArrB=1

x=-1to24=2ArArrA=12

rArrint(x^3+25)/(x^2+4x+3)dx=int12/(x+1)dx+int1/(x+3)dx

=12ln|x+1|+ln|x+3|+c

Cem Sentin
Jul 9, 2017

int(x^3+25)/(x^2+4x+3)dx

=int(x-4)dx+int(13x+27)/(x^2+4x+3)dx

=x^2/2-4x+int(13x+27)/((x+1)*(x+3))dx

Now I decomposed (13x+27)/((x+1)*(x+3))=A/(x+1)+B/(x+3) fraction

A*(x+3)+B*(x+1)=13x+27

(A+B)*x+3A+B=13x+27

After equating coefficients, I found A+B=13 and 3A+B=27 equations,

After solving them simultaneously, A=7 and B=6

Thus,

int(x^3+25)/(x^2+4x+3)dx

=x^2/2-4x+int(13x+27)/((x+1)*(x+3))dx

=x^2/2-4x+int7/(x+1)dx+int6/(x+3)dx

=x^2/2-4x+7Ln(x+1)+6Ln(x+3)+C

Explanation:

1) I took long division

2) I decomposed second integral into basic fractions

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