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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?

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Nothing. Note that $\frac{1}{30}$ is a common factor in the WA answer... $\frac{1}{30}*9=\frac{3}{10}$ ;) –  N. S. Jan 23 '12 at 3:23
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P.S. +1 for including all the details, I just wish my students would solve these problems so neatly ;) –  N. S. Jan 23 '12 at 3:25
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1 Answer

up vote 2 down vote accepted

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

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

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Wow... I can't believe I did not see that at the end. Well, I feel silly. –  Samuel Reid Jan 23 '12 at 3:24
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