1. ## Integrating fractions, help

$\int \frac{2x}{3-x} dx$

I have tired two methods, which both give me wrong answers, here my working , if anyone could tell me where I am going wrong, it would be greatly appreciated.

method 1: using a substitution:

$u = 3-x$

$x = 3-u$

$dx = -du$

$\int - \frac{2(3-u)}{u} du$

$-\int \frac{6-2u}{u} du$ = $-\int \frac{6}{u}-2 \\\ du$ = $- 6ln (3-x) + 2(3-x) + c$

so the answer is $6 - 2x -6ln(3-x) +c$

This is the wrong answer, but I can't see what I have dont wrong, if you anyone could kindly point out.

Thank you!

2. Originally Posted by Tweety
$\int \frac{2x}{3-x} dx$

I have tired two methods, which both give me wrong answers, here my working , if anyone could tell me where I am going wrong, it would be greatly appreciated.

method 1: using a substitution:

$u = 3-x$

$x = 3-u$

$dx = -du$

$\int - \frac{2(3-u)}{u} du$

$-\int \frac{6-2u}{u} du$ = $-\int \frac{6}{u}-2 \\\ du$ = $- 6ln (3-x) + 2(3-x) + c$

so the answer is $6 - 2x -6ln(3-x) +c$

This is the wrong answer, but I can't see what I have dont wrong, if you anyone could kindly point out.

Thank you!
whenever the degree of the numerator is $\ge$ degree of the denominator, perform long division ...

$\frac{2x}{3-x} = -2 + \frac{6}{3-x}$

now integrate.

3. Originally Posted by skeeter
whenever the degree of the numerator is $\ge$ degree of the denominator, perform long division ...

$\frac{2x}{3-x} = -2 + \frac{6}{3-x}$

now integrate.
thanks,

Can you also tell me weather this method is valid.

$\int \frac{4}{2} \times \frac{2}{4} \times \frac{2x}{3-x}$ which is multiplying by one,

$\frac{4}{2} \int \frac{4x}{-4x+12}$

= $\frac{4}{2}\int \frac{4x+12-12}{-4x+12}$

= $\frac{4}{2}\int \frac{4x+12}{-4x+12} - \frac{12}{-4x+12}$ =

$\frac{4}{2}\int -1 -\frac{12}{-4x+12}$

$\frac{4}{2} \int -1 - \frac{3}{3-x}$

which integrates to $\frac{4}{2}( -x - 3ln(3-x) )$

$-2x - 6ln(3-x) + c$

This is the correct answer, however I am unsure about weather I am able to multiply by 4/2 and 2/4 and than take 4/2 out of the integral?

4. of course you as long as you are not changing the overall function to integrate. Also $\int_a^b{cf(x)dx} = c\int_a^b{f(x)dx}$ for any constant c.