Question

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

Answer 1

`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`

Answer 2

`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