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

Apr 8, 2017

$I = - \frac{9 x + 11}{3 {\left(x + 2\right)}^{3}} + c ' .$

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:

$I = \int \frac{6 x + 5}{x + 2} ^ 4 \mathrm{dx} = \int \frac{\left(6 x + 12\right) - 7}{x + 2} ^ 4 \mathrm{dx} ,$

$= \int \frac{6 \left(x + 2\right)}{x + 2} ^ 4 \mathrm{dx} - 7 \int \frac{1}{x + 2} ^ 4 \mathrm{dx} ,$

$= 6 \int {\left(x + 2\right)}^{-} 3 \mathrm{dx} - 7 \int {\left(x + 2\right)}^{-} 4 \mathrm{dx} .$

We know that,

$\int f \left(x\right) \mathrm{dx} = F \left(x\right) + C \Rightarrow \int f \left(a x + b\right) \mathrm{dx} = \frac{1}{a} F \left(a x + b\right) + C , a \ne 0.$

Since $\int {x}^{-} 3 \mathrm{dx} = {x}^{- 3 + 1} / \left(- 3 + 1\right) = - \frac{1}{2 {x}^{2}} + c ,$ we have,

$I = 6 \left\{- \frac{1}{2 {\left(x + 2\right)}^{2}}\right\} - 7 \left\{{\left(x + 2\right)}^{- 4 + 1} / \left(- 4 + 1\right)\right\} ,$

$= \frac{7}{3 {\left(x + 2\right)}^{3}} - \frac{3}{x + 2} ^ 2 = \frac{7 - 9 \left(x + 2\right)}{3 {\left(x + 2\right)}^{3}} .$

$\therefore I = - \frac{9 x + 11}{3 {\left(x + 2\right)}^{3}} + c ' .$

Enjoy Maths.!