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

Question

1569 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-30T18:53:03

$int(6x+5)/(x+2)^4 dx = int (6/(x+2)^3-7/(x+2)^4)dx = -2/(x+2)^2+7/(3(x+2)^3) + C$

$(6x+5)/(x+2)^4 = A/((x+2)) + B/(x+2)^2 +C/(x+2)^3 +D/(x+2)^4$

$A(x+2)^3 + B(x+2)^2 +C(x+2) + D = 6x+5$

$Ax^3 = 0x^3 rArr A=0$

$Bx^2 = 0x^2 rArr B=0$

$Cx = 6x rArr C=6$

$2C+D=5 and C=6 rArr D=-7$

Now evaluate $int(6(x+2)^-3 - 7(x+2)^-4) dx$ to get

$-2/(x+2)^2+7/(3(x+2)^3) + C$

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