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
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
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.
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?
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.
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?
As I told you: $\displaystyle \frac{a^{m}}{a^{n}} = a^{m-n}$yes i know that. although i dont know WHY
Applying this to your case: $\displaystyle \frac{x^{n+1}}{x^{1}} = x^{n+1-1} = x^{n}$
--------------
$\displaystyle \int x^{n} \ln (ax) = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1}\int \frac{x^{n+1}}{x} dx$
$\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1} \int x^{n} dx$
$\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1} \cdot \frac{x^{n+1}}{n+1} + c$
$\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{x^{n+1}}{(n+1)^{2}} + c$
which agrees with your answer.