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}
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$\begingroup$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.
$\endgroup$ 2 $\begingroup$Use common denominators: $$\frac{\ln x\cdot x^{n+1}}{n+1}=\frac{(n+1)\cdot\ln x\cdot x^{n+1}}{(n+1)^2} $$
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