Use Integration by Parts to prove that $\int x^{n}\ln{x}\ dx=\frac{x^{n+1}}{(n+1)^{2}}\left[-1+(n+1)\ln{x}\right]+c$ I've gotten most of the way, but I can't see how I can transform my answer into the form in the assignment:
Use Integration by Parts to prove that $\displaystyle\int x^{n}\ln{x}\ dx=\frac{x^{n+1}}{(n+1)^{2}}\left[-1+(n+1)\ln{x}\right]+c$
\begin{align}
\int x^n\ln{x}\ dx&=\frac{\ln{x}\cdot x^{n+1}}{n+1}-\int\frac{x^{n+1}}{n+1}\cdot\frac{1}{x}\ dx\\
&=\frac{\ln{x}\cdot x^{n+1}}{n+1}-\frac{1}{n+1}\int x^n\ dx\\
&=\frac{\ln{x}\cdot x^{n+1}}{n+1}-\frac{1}{n+1}\left[\frac{x^{n+1}}{n+1}\right]+C\\
&=\frac{x^{n+1}}{(n+1)^2}+\dots
\end{align}

Your work is fine. We just need to algebraically "manipulate" the result a bit to get the answer to match the given equality (as stated in your title): 
Starting with your second-to-last line, we find a common denominator, $(n+1)^2$, and then factor out the common factor, which is the term you list in your last line, but is a factor (which multiplies over a sum/difference). So we have the common factor $\;\dfrac{x^{n+1}}{(n+1)^2}\times\Big[\cdots\Big]$:
$$
\begin{align}
\int x^n\ln{x}\ dx
& = \frac{\ln{x}\cdot x^{n+1}}{n+1}-\frac{1}{n+1}\left[\frac{x^{n+1}}{n+1}\right]+C \\ \\
& = \frac{(n+1)\ln x\cdot x^{n+1}}{(n+1)^2} - \frac{x^{n+1}}{(n+ 1)^2} + C \\ \\
& = \frac{x^{n+1}}{(n+1)^2}\Big[\left((n+1)\cdot \ln x\right) - 1\Big] + C \\ \\ 
& = \frac{x^{n+1}}{(n+1)^2}\Big[-1 + (n+1)\cdot \ln x\Big] + C \\ \\ 
\end{align}
$$
...which is now in the desired form.
