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How do I integrate this?

Apologies for the quality of the imageequation.jpg
Do you know how to do U-substitution?

Make U = x+1 and go from there.
Original post by Ash8991
Apologies for the quality of the imageequation.jpg


If this is the M3 question from 2008, you have separated the variables incorrectly. Your expression = v = dx/dt.

You therefore need to integrate the reciprocal of your expression.
Original post by coupdetat$
Do you know how to do U-substitution?

Make U = x+1 and go from there.


Original post by ChristopherHuey
Use u-substitution. Substitute u = x +1 and du = dx and work it through.

Reply with your attempt so we can check it for you.


Did you try it?
Reply 4
Avatar for Ash8991
Ash8991
OP
15
Original post by tiny hobbit
Did you try it?


Hey, thanks for responding. Before I checked this chat again, I did end up separating the variables and ended up getting:

(x+1)sqrt(6x(x+2))
------------------------ dx = dt
6x(x+2)

And with the substitution u = x + 1, I now have:

(u)sqrt(6(u-1)(u+1))
-------------------------- du = dt
6(u-1)(u+1)

Am I going in the right direction here? And yeah, this was from the 2008 M3 paper :smile:
Reply 5
Avatar for RichE
RichE
15
Original post by Ash8991
Hey, thanks for responding. Before I checked this chat again, I did end up separating the variables and ended up getting:

(x+1)sqrt(6x(x+2))
------------------------ dx = dt
6x(x+2)

And with the substitution u = x + 1, I now have:

(u)sqrt(6(u-1)(u+1))
-------------------------- du = dt
6(u-1)(u+1)

Am I going in the right direction here? And yeah, this was from the 2008 M3 paper :smile:


If your working is right to this point, you might try a trigonometric substitution to appropriately deal with the sqrt(u^2-1) term.
Reply 6
Avatar for Zacken
Zacken
22
Actually, your integral simplifies to
Unparseable latex formula:

\displaytyle \int \frac{u}{\sqrt{6}\sqrt{u^2 - 1}} \, \mathrm{d}u

- which doesn't need trig per se, it's either done instantly by recognition or use y=u2−1y = u^2 - 1 or y=x2y = x^2
Reply 7
Avatar for Ash8991
Ash8991
OP
15
Original post by Zacken
Actually, your integral simplifies to
Unparseable latex formula:

\displaytyle \int \frac{u}{\sqrt{6}\sqrt{u^2 - 1}} \, \mathrm{d}u

- which doesn't need trig per se, it's either done instantly by recognition or use y=u2−1y = u^2 - 1 or y=x2y = x^2


Thanks! I figured out where I went wrong. I tried that "y = u^2 - 1" substitution and then ended up getting the right answer. So thanks again! God, I hope a question like this doesn't come up in the exam...xD
Reply 8
Avatar for Zacken
Zacken
22
Original post by Ash8991
Thanks! I figured out where I went wrong. I tried that "y = u^2 - 1" substitution and then ended up getting the right answer. So thanks again! God, I hope a question like this doesn't come up in the exam...xD


Awesome, no worries.
Original post by Ash8991
Thanks! I figured out where I went wrong. I tried that "y = u^2 - 1" substitution and then ended up getting the right answer. So thanks again! God, I hope a question like this doesn't come up in the exam...xD


I thought at the time that this question was pushing the limits of what was reasonable based on C4 integration.

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