need help solving this integral problem

Question

el_arrow · Answer

$$\int\limits_{?}^{?} x(2x+5)^8$$

anonymous · Answer

integration by parts.

el_arrow · Answer

by subtitution

anonymous · Answer

whatever you want

el_arrow · Answer

i know the u= 2x+5 and du= 2dx but i dont know how to solve the problem

el_arrow · Answer

could you help me?

el_arrow · Answer

my book says that the answer is 1/40 (2x+5)^10 - 5/36 (2x+5)^9 + C but i have no idea how it got that

anonymous · Answer

I am not good enough to do it....

anonymous · Answer

I have to abandon you.

el_arrow · Answer

oh man

el_arrow · Answer

@zepdrix could you help me a moment please

zepdrix · Answer

sup broski? :U

el_arrow · Answer

hello zepdrix do you know hwo to do integrals by substituting?

zepdrix · Answer

$$\Large\rm \int\limits x(2x+5)^8 dx$$Yah a substitution will work out nicely! :)

el_arrow · Answer

yeah i got the u and du i just dont know how to work it out

zepdrix · Answer

$$\Large\rm u=2x+5$$$$\Large\rm \frac{1}{2}du=dx$$From that first piece of information, we can solve for x,$$\Large\rm \frac{1}{2}(u-5)=x$$

zepdrix · Answer

And then we plug these three pieces in.

el_arrow · Answer

okay

el_arrow · Answer

what do you do after that

zepdrix · Answer

$$\Large\rm \int\limits\limits \left[x\right](2x+5)^8 \left(dx\right)\quad=\quad \int\limits\limits \left[\frac{1}{2}(u-5)\right](u)^8 \left(\frac{1}{2}du\right)$$So uhhh, plugging everything in gives us something like this, yes?

el_arrow · Answer

i think so

zepdrix · Answer

The reason we did this is because it breaks up that 8th degree binomial.
From here we can distribute the u^8 to each term in the brackets.
Oh let's also pull the fractions out front.$$\Large\rm =\frac{1}{4}\int\limits u^9-5u^8~du$$

el_arrow · Answer

okay

el_arrow · Answer

and then we find the derivative of each right?

zepdrix · Answer

Yes.
If you can get it up to this point, it's easy peasy from here.
Just power rule each term.

Getting it up to this point can be a little tricky though :)
Look back at those steps some more and let me know if any were too confusing.

Oh, and don't forget to undo your substitution after you integrate.

el_arrow · Answer

how do you undo the substitution

zepdrix · Answer

Well, like, our first term would give us:$$\Large\rm \frac{1}{10}u^{10}$$But our u was 2x+5.
So we plug that back in.$$\Large\rm \frac{1}{10}(2x+5)^{10}$$Keep in mind I didn't apply the 1/4 that was in front, this was just to give you a boost in the right direction :U

el_arrow · Answer

oh okay thank you so much you make it easier to understand

zepdrix · Answer

cool c: