Partial Fractions - Integration

[emmaerin]

Homework Statement

Evaluate the integral. (Remember to use ln |u| where appropriate. Use C for the constant of integration.)

$\int \frac {5x^2 - 20x +45}{(2x+1)(x-2)^2}\, dx$

Homework Equations

$5x^2 - 20x +45$ = $5 (x^2 -4x +9)$

The Attempt at a Solution

I'm able to come up with an answer, but it's not correct. I'd really appreciate it if someone could tell me where I'm going wrong. Thanks! (Also this is my first attempt at LaTeX, so forgive me if it's not formatted properly!)

First to solve for the partial fractions:
$\frac {A}{2x+1} + \frac {B}{x-2} + \frac {C}{(x-2)^2}$

$x^2 - 4x +9 = A (x-2)^2 + B (2x+1) (x-2) + C (2x+1)$
$x^2 - 4x + 9 = A (x^2 - 4x + 4) + B (2x^2 - 3x - 2) + C (2x+1)$

$x^2 = x^2 (A+2B)$
$-4x = x (-4A - 3B + 2C)$
$9 = 4A -2B +C$

Therefore,
$C = 9-4A +2B$
$-4x = x ( -4A - 3B + 2(9-4A+2B))$
$-4 = -12A + B +18$
$-22 = -12A + B$

and

$1 = A + 2B$
$-22 = -22A - 44B$

Gives me,

$-12A + B = -22A - 44B$
$A = \frac {-9}{2} B$
$1 = \frac {-9}{2}B + 2B$

*$\mathbf B = \frac{-2}{5}$

$1 = A + 2B$

*$\mathbf A = \frac{9}{5}$

$C = 9 - 4A + 2B$

*$\mathbf C = 1$

Substituting in A, B, and C gives me:

$\int \frac {\frac{9}{5}}{2x+1}\,dx$ + $\int \frac {\frac{-2}{5}}{x-2} \,dx$ + $\int \frac {1}{(x-2)^2}\,dx$

$= \frac{9}{5} \int \frac {1}{2x+1}\,dx$ $- \frac{2}{5} \int \frac {1}{x-2}\,dx$ $+ \int \frac {1}{(x-2)^2} \, dx$For the first integral, $= \frac{9}{5} \int \frac {1}{2x+1}\,dx$
$u = 2x+1, du = 2 dx$
$\frac {9}{10} \int \frac {1}{u}\, du$
$= \frac {9}{10} ln |2x+1|$For the second integral, $\int \frac {\frac{-2}{5}}{x-2} \,dx$
$\frac{-2}{5} \int \frac {1}{x-2} \, dx$
$= \frac{-2}{5} ln |x-2|$ For the third integral, $\int \frac {1}{(x-2)^2} \, dx$
$\int (x-2)^{-2} \, dx$
$= \frac {-1}{x-2}$This gives me the following answer, but it's wrong :( I know this is a long problem, so any help would be appreciated!

$\frac {9}{10} ln |2x+1|$ + $\frac{-2}{5} ln |x-2|$ + $\frac {-1}{x-2}$ +C

 

[HakimPhilo]

Revise your calculations of A, B and C, when you decompose your fraction, that's where the error is.

 

[gopher_p]

A) You factored out a 5 from the numerator at the beginning of the problem, and then it disappeared.
B) Your answer involves logs, and log laws might have been used to make the book's (or whatever it is you're comparing your answer to) answer look different from yours.

 

[emmaerin]

A) You factored out a 5 from the numerator at the beginning of the problem, and then it disappeared.

I completely forgot to bring down the 5 - that was the whole problem! Such a silly mistake, thanks for pointing it out!