This integral turns out to be easier than you might think. First of all, note that we have to use integration by parts. Integration by parts is defined as follows:

$$ \int uv'\, dx = uv - \int u'v\, dx $$

When computing $ \int x^{n} \log(x)\, dx $, we will want to differentiate $x^n$ and integrate $\log(x)$. Therefore, we choose $u = x^n$ and $v' = \log(x)$. First of all, note that the integral of $\log(x)$ is $x\log(x) - x$, and it follows that $v = x\log(x) - x$. By differentiation, we find that $u' = nx^{n-1}$.

Using integration by parts, we have

$$ \int x^{n} \log(x)\, dx = x^n (x\log(x) - x) - \int (nx^{n-1})(x\log(x) - x)\,dx. $$

Note that the second term on the right-hand side can be simplified tremendously! We factor:

$$ \int x^{n} \log(x)\, dx = x^{n+1} \log(x) - x^{n+1} - \int nx^{n}\log(x) - nx^{n}\,dx. $$

We bring $n$ in front of the rightmost integral and we split it:

$$ \underbrace{\int x^{n} \log(x)\, dx}_{\text{These two integrals...}} = x^{n+1} \log(x) - x^{n+1} - n\underbrace{\int x^{n}\log(x)\,dx}_{\text{... are identical}} + n\int x^{n}\,dx. $$

We add $n\int x^{n}\log(x)\,dx$ on both sides and we compute the rightmost integral, which is trivial:

$$ \int x^{n} \log(x)\, dx + n\int x^{n}\log(x)\,dx = x^{n+1} \log(x) - x^{n+1} + \frac{n}{n+1} x^{n+1}. $$

We factor on both sides:

$$ (n+1)\int x^{n} \log(x)\, dx = x^{n+1} \left(\log(x) - 1 + \frac{n}{n+1}\right) $$

Finally, we have:

$$ \int x^{n} \log(x)\, dx = \frac{x^{n+1}}{n+1} \left(\log(x) - \frac{1}{n+1}\right) + C $$

Never forget the constant!