Let's take this as an example.
$\int (x^{15}\ln x)dx $
Is there a way to solve it someway clever? Using integration by parts 15 times would be wearisome...
Let's take this as an example.
$\int (x^{15}\ln x)dx $
Is there a way to solve it someway clever? Using integration by parts 15 times would be wearisome...
Integrating by parts once is enough (but integrating by parts 15 times, as suggested in the original posting, is not enough, as you'll find if you actually do it): \begin{align} \int x^{15} \ln x\,dx & = \int \Big(\ln x\Big)\Big(x^{15} \, dx\Big) \\[10pt] & = \int u\,dv \\[10pt] & = uv - \int v\,du \\[10pt] & = \Big(\ln x\Big) \Big( \frac{x^{16}}{16} \Big) - \int \Big(\frac{x^{16}}{16}\Big) \Big(\frac{dx}{x} \Big) \\[10pt] & = \frac{x^{16}\ln x}{16} - \frac{1}{16} \int x^{15}\,dx \\[10pt] & = \frac{x^{16}\ln x}{16} - \frac{1}{16}\cdot\frac{x^{16}}{16} + C. \end{align}
Hint: $\int x^{15}\ln x dx=\frac1{16}\int(x^{16})^{\prime}\ln x dx$ Then you only need $1$ integration by parts.
Set $\ln(x) = t$. Hence, we get $x = e^t \implies dx = e^t dt$. Hence, \begin{align} I_n & = \int x^n \log(x) dx = \int e^{nt} t e^t dt = \int t e^{(n+1)t} dt = \dfrac1{n+1} \int t d(e^{(n+1)t})\\ & = \dfrac1{n+1} \left( te^{(n+1)t} - \int e^{(n+1)t} dt\right) + c = \dfrac1{n+1} \left(te^{(n+1)t} - \dfrac{e^{(n+1)t}}{n+1} \right)+c\\ & = \dfrac1{n+1} \left(\ln(x) - \dfrac1{n+1} \right)x^{n+1} + c \end{align} Set $n=15$, to get what you want.
You can use integration by parts and get a formula for the general case.
Note that, first, $\frac{d}{dx}(x \log{x} - x ) = \log{x}$
Now, for the general case, integrate by parts:
$\int dx \: x^k \log{x} = x^{k+1} (\log{x} - 1) - k \int dx \: x^k \log{x} + k \int dx \: x^k$
or,
$ (k+1) \int dx \: x^k \log{x} = x^{k+1} (\log{x} - 1) + \frac{k}{k+1} x^{k+1}$
Now just plug in $k=15$ and you are done.
Intgration by parts is, in this case, not wearisome. Let $du=x^{15}\,dx$, and let $v=\ln x$. (Yes, kind of backwards.)
Then we can take $u=\dfrac{x^{16}}{16}$. Also, $dv=\dfrac{1}{x}\,dx$.
So at the next step we are finding $\displaystyle\int \dfrac{x^{16}}{16}\dfrac{1}{x}\,dx$, that is, $\displaystyle\int\dfrac{x^{15}}{16}\,dx$. This one is immediate.
As noted, the high $x$ exponent does not complicate things too much as $\ln x$ gives something workable once you take the derivative while integrating it will not eliminate $\ln x$ from your integral. If you want it to look a little less formidable though or already have $\int x\ln xdx$ memorized, proper substitution will clean it up.
$u=x^8,du=8x^7dx$
$\int x^{15}\ln xdx=\frac18\int u\ln u^\frac18du=\frac1{64}\int u\ln udu$
Integration by parts is not needed. Note that $\dfrac{d}{dt} x^t = x^t \ln(x)$. Now $\displaystyle\int x^t \ dx = \dfrac{x^{t+1}}{t+1} + c$ so $\int x^t \ln(x)\ dt = \dfrac{d}{dt} \dfrac{x^{t+1}}{t+1} = \dfrac{(t+1) x^{t+1} \ln(x) - x^{t+1}}{(t+1)^2} + c$