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December 18, 2014

December 18, 2014

Posted by **Eric** on Wednesday, May 15, 2013 at 10:51am.

1/(t^2+t+1) dt

how do i even start. do i integrate and then plug in a and b? plz help.

- integral -
**Steve**, Wednesday, May 15, 2013 at 11:19amdy/dx = 1/(x^2+x+1) * 3

- integral -
**Eric**, Wednesday, May 15, 2013 at 11:30amcan you explain how you got this. thanks

- integral -
**Steve**, Wednesday, May 15, 2013 at 11:35amexcuse me? Didn't we just go through this about differentiating under the integral sign?

- integral -
**Eric**, Wednesday, May 15, 2013 at 11:38ambut i dont get how you can just replace the variable t with x so easily. i dont get how you did that. and how does the 3 come into the picture for dy/dx.

- integral -
**Steve**, Wednesday, May 15, 2013 at 12:04pmIn general (see the wikipedia article for proof),

∫[a(x),b(x)] f(t) dt

= f(b(x)) db/dx - f(a(x)) da/dx

Since we have a(x) = 1, da/dx = 0 and we are left with

∫[a(x),b(x)] f(t) dt

= f(b(x)) db/dx

b(x) = 3x, so db/dx = 3

So, oops. I made a mistake. It should be

dy/dx = 1/((3x)^2+3x+1) * 3

Good catch! :-)

- integral -
**Eric**, Wednesday, May 15, 2013 at 12:12pmThank you very much! I really didn't understand the form but now I do. :)

Also, would d2y/dx2 be:

-9(6x+1)/((9x^2+3x+1)^2)

I just want to make sure that d2y/dx2 is asking for the second derivative?

- integral -
**Steve**, Wednesday, May 15, 2013 at 12:42pmhaven't checked your work, but yes, they want y"

- integral -
**Eric**, Wednesday, May 15, 2013 at 1:49pmOh okay thank you!

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