# 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}

• Have you tried just differentiating the RHS? – muzzlator Mar 3 '13 at 23:01
• Why don't you differentiate the right side? It's pretty simple. – MyUserIsThis Mar 3 '13 at 23:02
• Sorry - I didn't copy the question word for word. I have to use IBP. I updated the question now. – agent154 Mar 3 '13 at 23:03

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.

• agent154: Are you okay with this? Do you see how to get $\dfrac{x^{n+1}}{(n+1)^2}\Big[-1 + (n+1)\cdot \ln x\Big] + C$ and not $\dfrac{x^{n+1}}{(n+1)^2}\Big[1 - (n+1)\cdot \ln x\Big] + C$ – Namaste Mar 4 '13 at 15:18
• Yes, I do see that. Thanks. – agent154 Mar 10 '13 at 0:10

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

• Is there a typo in the question I wonder? The best I can come up with is 1-(n+1)ln(x), whereas the question says it should be −1+(n+1)ln(x) – agent154 Mar 3 '13 at 23:20
• @agent154 It looks like you should get the given answer from the work above. If you're off by a sign, the mistake comes after the work you've completed above. – Mike Mar 3 '13 at 23:33
• @agent154: I still can't see your problem. What you have calculated up there, is just $\left((n+1)\cdot\ln x-1\right)\cdot\frac{x^{n+1}}{(n+1)^2}$ as requested. – Berci Mar 4 '13 at 10:44
• I found it - I just mixed up my signs after that up there.. I forgot that there should be a negative sign in front of the $\frac{x^{n+1}}{(n+1)^2}$, which allows me to get the $-1+(n+1)\ln(x)$ – agent154 Mar 4 '13 at 14:43