# Thread: Integrating fraction w/ powers, can't work out last term coefficient: int x/(x+3)^2

1. ## Integrating fraction w/ powers, can't work out last term coefficient: int x/(x+3)^2

Hi

I did a calculation on an integral but i get the wrong answer, but the correct answer is elusive to me in terms of how it got the final term.

This is the workings i did:

. . . . .$\displaystyle \int\, \left(\dfrac{x}{x\, +\, 3}\right)^2\, dx\, =\, \int\, \dfrac{x^2}{(x\, +\, 3)^2}\, dx$

Long division gives us:

. . . . .$\displaystyle \int\, \left(1\, -\, \dfrac{6x\, -\, 9}{(x\, +\, 3)^2}\right)\, dx$

Partial-fraction decomposition gives us:

. . . . .$\displaystyle \dfrac{6x\, -\, 9}{(x\, +\, 3)^2}\, =\, \dfrac{A}{x\, +\, 3}\, +\, \dfrac{B}{(x\, +\, 3)^2}$

. . . . .$\displaystyle 6x\, -\, 9\, =\, A(x\, +\, 3)\, +\, B$

. . . . .$\displaystyle 6x\, -\, 9\, =\, Ax\, +\, (3A\, +\, B)$

. . . . .$\displaystyle 6\, =\, A$

. . . . .$\displaystyle -9\, =\, 3A\, +\, B\, =\, 18\, +\, B$

. . . . .$\displaystyle -27\, =\, B$

So the integral becomes:

. . . . .$\displaystyle \int\, \left(1\, -\, \dfrac{6}{x\, +\, 3}\, -\, \dfrac{27}{(x\, +\, 3)^2}\right)\, dx$

. . . . . . . .$\displaystyle=\, \int\, 1\, dx\, -\, 6\, \int\, \dfrac{1}{x\, +\, 3}\, dx\, -\, 27\, \int\, \dfrac{1}{(x\, +\, 3)^2}\, dx$

. . . . . . . .$\displaystyle =\, x\, -\, 6\, \ln|x\, +\, 3|\, -\, \dfrac{27}{x\, +\, 3}\, +\, C$

. . . . .$\displaystyle \large{\mathbf{\color{#ff8c00}{ \mathit{x}\, -\, 6\, \ln\big|\mathit{x}\, +\, 3\big|\, -\, \dfrac{9}{\mathit{x}\, +\, 3}\, +\, C }}}$

I cannot see how they get -9 over x+3 but wolfram says its -9 also, so i am clearly making an error here but i cannot see it.

Hope some one can see my error.

2. Hi,

The error is at the beginning. The remainder of the division is indeed $-6x-9 = -(6x+9)$. The integrand should be rewritten as:

$\displaystyle1-\frac{6x+9}{(x+3)^2}$

Note that, independently of this error, you made another sign error in the last line of your calculation; you should have written:

$\displaystyle-27\int\frac{1}{(x+3)^2}=+\frac{27}{x+3}$

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