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For practice, I am integrating,

$\int \frac{x}{3x^2 + 8x -3}dx$

So, I can then factor it as,

$\int \frac{x}{(3x-1)(x+3)}dx$

By partial fractions, I decompose

$\frac{x}{(3x-1)(x+3)}= \frac{A}{3x-1} + \frac{B}{x+3}$

For finding $A$, I multiply both sides by $3x-1$, which gives

$\frac{x(3x-1)}{(3x-1)(x+3)} = \frac{A(3x-1)}{3x-1} + \frac{B(3x-1)}{x+3}$

So, we have that

$\frac{x}{(x+3)} = A + \frac{B(3x-1)}{x+3}$

Letting $3x-1=0$, we have that $x=\frac{1}{3}$, so then

$\frac{\frac{1}{3}}{(\frac{1}{3}+3)} = A$

Thus, we have that $A=\frac{1}{10}$. For determining $B$, we then multiply both sides by $x+3$ and receive, as a similar process to the previous,

$\frac{x(x+3)}{(3x-1)(x+3)} = \frac{A(x+3)}{3x-1} + \frac{B(x+3)}{x+3}$

Then,

$\frac{x}{3x-1} = \frac{A(x+3)}{3x-1} + B$

So, if we let $x+3=0$, we then have that $x=-3$ and so,

$\frac{-3}{3(-3)-1}=B$

So, we then have that $B=\frac{3}{10}$. Thus, our original integral can then be written as,

$\int \frac{x}{(3x-1)(x+3)}dx = \int \frac{1}{10(3x-1)} + \frac{3}{10(x+3)} dx$

We can, by splitting up the integral find,

$\int \frac{x}{(3x-1)(x+3)}dx = \frac{1}{10} \int \frac{1}{3x-1} dx + \frac{3}{10} \int \frac{1}{x+3} dx$

Thus, we conclude that,

$\int \frac{x}{3x^2 + 8x -3}dx = \frac{\ln|3x-1|}{30} + \frac{3 \ln|x+3|}{10} + C$

Wolframalpha shows that, the answer is:

$\frac{1}{30}(\ln(1-3x)+ 9 \ln(3+x)) +C$

What am I doing wrong, did I miss a negative sign somewhere?

  • 1
    P.S. +1 for including all the details, I just wish my students would solve these problems so neatly ;)2012-01-23

1 Answers 1

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Wolfram forgot the absolute value signs. Further both answers are the same.

$\frac{9}{30}=\frac{3}{10}$

  • 0
    Wow... I can't believe I did not see that at the end. Well, I feel silly.2012-01-23