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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