The Student Room Group

Integration by substitution problem

Hi

I was wondering if anyone could help me with this problem which I can't work out how to solve.

4x(2x+1)5dx\int 4x(2x+1)^5dx u=2x+1 u = 2x+1

This is what I have done:

du=2dx du = 2dx
dx=12dudx = \frac{1}{2} du

4x(u)512du\int 4x (u)^5 \frac{1}{2}du

124x(u)5du\frac{1}{2}\int4x (u)^5 du

But I am not sure how to get rid of the 4x and proceed from here. Any help would be much appreciated!
Original post by jonnburton
Hi

I was wondering if anyone could help me with this problem which I can't work out how to solve.

4x(2x+1)5dx\int 4x(2x+1)^5dx u=2x+1 u = 2x+1

This is what I have done:

du=2dx du = 2dx
dx=12dudx = \frac{1}{2} du

4x(u)512du\int 4x (u)^5 \frac{1}{2}du

124x(u)5du\frac{1}{2}\int4x (u)^5 du

But I am not sure how to get rid of the 4x and proceed from here. Any help would be much appreciated!


u = 2x+1
Therefore...
Reply 2
Original post by jonnburton
Hi

I was wondering if anyone could help me with this problem which I can't work out how to solve.

4x(2x+1)5dx\int 4x(2x+1)^5dx u=2x+1 u = 2x+1

This is what I have done:

du=2dx du = 2dx
dx=12dudx = \frac{1}{2} du

4x(u)512du\int 4x (u)^5 \frac{1}{2}du

124x(u)5du\frac{1}{2}\int4x (u)^5 du

But I am not sure how to get rid of the 4x and proceed from here. Any help would be much appreciated!


What is 4x4x in terms of uu?
Reply 3
So 4x would be 2u -2?
Original post by jonnburton
So 4x would be 2u -2?


Or 2(u-1) which may help further. :smile:
Reply 5
OK, thanks m4ths!

Taking this further, I get:

122(u1)(u)5du \frac{1}{2}\int2(u-1)* (u)^5 du

Taking out the 2 gives:
(u1)(u)5du \int(u-1)* (u)^5 du

and then integrating this:

12(u1)2u66\frac{1}{2}(u-1)^2 * \frac{u^6}{6}


I think I have gone wrong with the integration of (u-1) somehow because according to the book's answer this is supposed to come out at u77\frac{u^7}{7} (before substituting the original value of u back in).

I can't see how what I've done is wrong though because I've followed the principle: (ax+b)ndx=1a(n+1)(ax+b)n+1\int(ax+b)^n dx = \frac{1}{a(n+1)}(ax+b)^{n+1}
Reply 6
Original post by jonnburton


(u1)(u)5du \int(u-1)* (u)^5 du

and then integrating this:

12(u1)2u66\frac{1}{2}(u-1)^2 * \frac{u^6}{6}



Oh dear

You cannot do that ... basic error

You should have done (u1)u5=u6u5(u-1)u^5 = u^6 - u^5 then integrated
Reply 7
Oh, thanks TenOfThem. It makes sense now.
Reply 8
Original post by jonnburton
Hi

I was wondering if anyone could help me with this problem which I can't work out how to solve.

4x(2x+1)5dx\int 4x(2x+1)^5dx u=2x+1 u = 2x+1

This is what I have done:

du=2dx du = 2dx
dx=12dudx = \frac{1}{2} du

4x(u)512du\int 4x (u)^5 \frac{1}{2}du

124x(u)5du\frac{1}{2}\int4x (u)^5 du

But I am not sure how to get rid of the 4x and proceed from here. Any help would be much appreciated!

Are you being asked to use substitution here? Personally I would just do it directly :smile: (sorry to type up a solution, but, as your question has been answered I thought I would include this as a bonus if you're interested)

2((2x+1)1)(2x+1)5 dx=2(2x+1)6(2x+1)5 dx[br]=17(2x+1)716(2x+1)6+C\displaystyle 2 \int ((2x+1)-1)(2x+1)^5 \ dx = 2 \int (2x+1)^6 - (2x+1)^5 \ dx [br]\displaystyle = \frac{1}{7} (2x+1)^7 - \frac{1}{6} (2x+1)^6 + \mathcal{C}
Reply 9
Thanks for that working Jkn; this is an exercise to practice substitution so I was supposed to do it that way.

I do have one question about your working, if you could answer it. How does [latex](2x+1)-1)[\latex] become (2x+1)6(2x+1)^6?

Thanks
Reply 10
[QUOTE="jonnburton;43062358"]Thanks for that working Jkn; this is an exercise to practice substitution so I was supposed to do it that way.

I do have one question about your working, if you could answer it. How does
Unparseable latex formula:

(2x+1)-1)[\latex] become [latex](2x+1)^6[/latex]?[br][br]Thanks



(2x+1)5((2x+1)1)=(2x+1)5(2x+1)(2x+1)5=(2x+1)6(2x+1)5(2x + 1)^{5}((2x + 1) - 1) = (2x + 1)^{5}(2x + 1) - (2x + 1)^{5} = (2x + 1)^{6} - (2x + 1)^{5}
Reply 11
I miss simple things too often..! Thanks a lot for showing that!

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