Integrals... yeah...

[Ducky Havok]

Well, on my final today for AP Calc. BC, I was able to solve all the integrals except this one... It's bugging me because I feel like I'm missing something obvious, and my teacher put them in order from easiest to hardest and this was somewhere toward the middle. The test is over so it doesn't matter now, I'm just curious as to how it should have been solved (without a calculator).

 

[math]\int{x^{5}\ln{(3x)}dx}[/math]

 

Edit: Corrected version of problem (I wrote it wrong, I'm a loser )

[math]\int x^{5}\ln(x+3)dx[/math]

[Tom Mattson]

I bet the thing you're missing is the fact that [imath]\ln(3x)=\ln(x)+\ln(3)[/imath]. If you split the natural log up that way you get one integral that can be done by the power rule and another one that can be done by parts.

[Ducky Havok]

Thanks, but that stills leaves [math]\int{x^5\ln(x)dx}[/math] which is pretty much the same problem. I hadn't thought about that or else I probably could have gotten partial credit, though. My thoughts were to turn it into a Maclaurin series and then integrate it (because that was one of the last things we went over), but I had some trouble with that. And after all that I just realized something. Since I never wrote the question down, I typed it wrong. I'm sorry. Here's the corrected version.

 

[math]\int{x^5\ln(x+3)}[/math]

 

Thanks for the help on the first one though, hopefully you can help on the real version now.

[Dave]

Surely integration by parts is the obvious approach?

 

[math]\int x^5\ln(x)dx = \tfrac{1}{6} x^6 \ln x - \tfrac{1}{6}\int x^5 dx[/math]

 

Edit: I should think a similar trick will work for the second integral as well.

[Tom Mattson]

Thanks' date=' but that stills leaves [math']\int{x^5\ln(x)dx}[/math] which is pretty much the same problem.

 

Yeah, but it's an easy problem. You do it by parts: Let [imath]u=\ln(x)[/imath] and [imath]dv=x^5dx[/imath].

 

Here's the corrected version.

 

[math]\int{x^5\ln(x+3)}[/math]

 

Same type of thing, you do it by parts. Let [imath]u=\ln(x+3)[/imath] and [imath]dv=x^5dx[/imath].

[Ducky Havok]

Okay, thanks Dave, your approach makes perfect sense, and I kind of forgot about that at the time (this is all self taught so I may have just overlooked it). But, that was for the first one which I think was a little easier because the 2nd part simplified.

Using Tom's method, I get [math]\frac{x^6\ln(x+3)}{6}-\frac{1}{6}\int{\frac{x^6}{x+3}dx}[/math]

And thus comes a new dilemma for me. Maybe I should just give up for the night and look at it again after sleep... that usually seems to help. Thanks for the help though. (I think I've become discouraged because I used an online integrator to solve it to check for an answer and I got something extremely long)

[Tom Mattson]

Using Tom's method' date=' I get [math']\frac{x^6\ln(x+3)}{6}-\frac{1}{6}\int{\frac{x^6}{x+3}dx}[/math]

And thus comes a new dilemma for me.

 

No, it's not a dilemma. It's trivial.

 

Do long division, as you would with any improper rational function. Then integrate the polynomial and the proper rational function and you're done.

[chuinhen]

\int x^5\ln(x+3)dx = \tfrac{1}{6} x^6 \ln x - \tfrac{1}{6}\int [x^5\int ln(x+3)dx] dx