1. ## Integration Help

hey there..
i have an assignment to do and dont really know how to do this question...

$\displaystyle \int x^n ln(ax) dx$ (n cant equal -1)

do i do it by substitution? parts???
where do i start..???
if someone could help would be great. thanks

2. Just directly apply integration by parts starting with:
$\displaystyle \begin{array}{lll} u = \ln (ax) & \quad & dv = x^{n} dx \\ du = \text{???} & \quad & v = \text{???}\end{array}$

Then use the fact that: $\displaystyle \int u dv = uv - \int v du$

3. what would du be in this case??? not very great with logs...

would it be 1/(ax) ???

4. You must've learned about the chain rule right?

$\displaystyle \left[f(g(x))\right]' = f'(g(x)) \cdot g'(x)$

Imagine $\displaystyle f(x) = \ln (x)$ and $\displaystyle g(x) = ax$
$\displaystyle \left[\ln (ax)\right]' = \frac{1}{ax} \cdot (ax)'$

etc. etc.

5. yes. i have

i got this far and now stuck again. dont know waht to do..

x^n ln(ax) dx

Let u = ln(ax)
du = 1/ax . a = 1/x
dv = x^n dx
v = x^(n+1) / (n+1)

udv = uv - vdu

ln(ax) . (x^(n+1)) / (n+1) - (x^(n+1)) / (n+1) . 1/x

how do i proceed now??

6. Originally Posted by b00yeah05
yes. i have

i got this far and now stuck again. dont know waht to do..

x^n ln(ax) dx

Let u = ln(ax)
du = 1/ax . a = 1/x
dv = x^n dx
v = x^(n+1) / (n+1)

udv = uv - vdu

ln(ax) . (x^(n+1)) / (n+1) - (x^(n+1)) / (n+1) . 1/x

how do i proceed now??
The integration by parts formula is $\displaystyle \int u \, dv = uv - \int v \, du$.

After making the appropriate substitutions, the integral on the right hand side should be blue sky all the way ......

7. sorry i knew the forumla just didnt know how to type it using the latex thing. i have multiplied them together and dont know how to proceed...

on the right hand side i get...

x^(n+1) / x(n+1)

i know that the x's cancel out but then how do i integrate the restt?? and how do i simplify the left hand side?

8. Ok so making your substitutions into the formula:
$\displaystyle \int x^{n} \ln (ax) dx = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \int \left(\frac{x^{n+1}}{n+1} \cdot \frac{1}{x} \right) dx$

$\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1}\int \frac{x^{n+1}}{x} dx$ (Factored out the constant $\displaystyle \frac{1}{n+1}$ )

Remember that: $\displaystyle \frac{a^{m}}{a^{n}} = a^{m-n}$

etc. etc.

9. sorry would it be possible for you to do the rest of the working and give an answer pl.ease so i can understand?? i have never come accross a question like this..i have really tried to do it but i cant get around it.

10. Originally Posted by b00yeah05
sorry would it be possible for you to do the rest of the working and give an answer pl.ease so i can understand?? i have never come accross a question like this..i have really tried to do it but i cant get around it.
Are you saying you can't find $\displaystyle \int \frac{x^{n+1}}{x} dx$?

11. yes...that and i dont know how to simplify the left hand side...

12. Do you realise that $\displaystyle \int \frac{x^{n+1}}{x} dx = \int x^n\, dx$ ?

13. yes i know that. although i dont know WHY...so i was just looking for the working for that...i cant seem to get around it. therefore because of the factor that was taken out before....would it just be (x^(n+1)) / (n+1)^2 ????

therefore the full answer would be....

ln(ax) . (x^(n+1)) / (n+1) - (x^(n+1))/ (n+1)^2

?? am i correcT?

14. yes i know that. although i dont know WHY
As I told you: $\displaystyle \frac{a^{m}}{a^{n}} = a^{m-n}$
Applying this to your case: $\displaystyle \frac{x^{n+1}}{x^{1}} = x^{n+1-1} = x^{n}$

--------------

$\displaystyle \int x^{n} \ln (ax) = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1}\int \frac{x^{n+1}}{x} dx$
$\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1} \int x^{n} dx$
$\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1} \cdot \frac{x^{n+1}}{n+1} + c$
$\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{x^{n+1}}{(n+1)^{2}} + c$