Question

How do you integrate `(6x+5) / (x+2) ^4` using partial fractions?

Answer 1

`I=-(9x+11)/(3(x+2)^3)+c'.`

Explanation:

I feel that there is no need to use the Method of Partial Fractions,

because, without it, the Problem can be easily worked out, as shown

below:

`I=int(6x+5)/(x+2)^4dx=int{(6x+12)-7}/(x+2)^4dx,`

`=int(6(x+2))/(x+2)^4dx-7int1/(x+2)^4dx,`

`=6int(x+2)^-3dx-7int(x+2)^-4dx.`

We know that,

`intf(x)dx=F(x)+C rArr intf(ax+b)dx=1/aF(ax+b)+C, ane0.`

Since `intx^-3dx=x^(-3+1)/(-3+1)=-1/(2x^2)+c,` we have,

`I=6{-1/(2(x+2)^2)}-7{(x+2)^(-4+1)/(-4+1)},`

`=7/(3(x+2)^3)-3/(x+2)^2={7-9(x+2)}/(3(x+2)^3).`

`:. I=-(9x+11)/(3(x+2)^3)+c'.`

Enjoy Maths.!