# Integration Help - Which Method?

1. Feb 4, 2014

### BOAS

Hello,

i'm having some trouble identifying the correct method to approach a problem. I don't have any working to show, so i'll explain why I think the methods i've considered are not correct.

1. The problem statement, all variables and given/known data

Integrate with respect to $x$ the following functions.

$\int \frac{x}{(3-x)^{7}} dx$

2. Relevant equations

3. The attempt at a solution

I think the method of changing the variable is not suitable here because I can't see a link between the derivative of the denominator and the numerator.

Similarly I don't think it is possible to use the method that identifies the numerator as the derivative of the denominator allowing you to say the integral is equal to ln|f(x)| + c

I'd really appreciate a nudge in the right direction!

Thank you,

BOAS

2. Feb 4, 2014

### ehild

Use the substitution u=3-x.

ehild

3. Feb 4, 2014

### BOAS

I'm struggling to make this work.

$\int \frac{x}{(3-x)^{7}} dx$

Let $u \equiv (3 - x)$

$\frac{du}{dx} = -1$

$... du \equiv ... -1dx$

$\int \frac{x}{(3-x)^{7}} dx = -x \int \frac{du}{u^{7}}$

(that seems wrong to me, I don't think we've ever been told it's ok to bring an x outside the integral sign)

I get;

$\int \frac{x}{(3-x)^{7}} dx = \frac{x}{6(3-x)^6} + c$

4. Feb 4, 2014

### phyzguy

You can't just pull out the x like you did. You have to replace the x in the numerator with x=3-u and leave it under the integral sign.

5. Feb 4, 2014

### Tanya Sharma

The item in red is wrong . You need to express 'x' present in the Nr also in terms of 'u' .After the substitution ,you should have $\int \frac{u-3}{u^7} du$

6. Feb 4, 2014

### Ray Vickson

Others have suggested a change of variables, but if you do straight integration by parts you can bypass that: just set $u = x, \: dv = dx/(3-x)^7 = (3-x)^{-7} \, dx$.

7. Feb 4, 2014

### statdad

Think again about your substitution: if $u = 3 -x$ then $x = 3 - u$. How will that help your integrand

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