# An Integral

• Jun 2nd 2008, 04:55 PM
zellster87
An Integral
It's been a while since I did integrals, and i'm having trouble with this problem.

√( (2t)^2 + (t^-2) + (4) ) And the integral goes from 1 to e

Is there an easier way I can simplify the stuff under the square root, in order to make this less messy? Or do I have to go straight to substitution? And if so, whats the best way to start? Thanks for your time.
• Jun 2nd 2008, 05:05 PM
Mathstud28
Quote:

Originally Posted by zellster87
It's been a while since I did integrals, and i'm having trouble with this problem.

√( (2t)^2 + (t^-2) + (4) ) And the integral goes from 1 to e

Is there an easier way I can simplify the stuff under the square root, in order to make this less messy? Or do I have to go straight to substitution? And if so, whats the best way to start? Thanks for your time.

$\displaystyle \int(2t)^2+t^{-2}+4dt$

simplifying to

$\displaystyle \int{4t^2+t^{-2}+4dt}=\frac{4}{3}t^3-t^{-1}+4t+C$
• Jun 2nd 2008, 05:05 PM
topsquark
Quote:

Originally Posted by zellster87
It's been a while since I did integrals, and i'm having trouble with this problem.

√( (2t)^2 + (t^-2) + (4) ) And the integral goes from 1 to e

Is there an easier way I can simplify the stuff under the square root, in order to make this less messy? Or do I have to go straight to substitution? And if so, whats the best way to start? Thanks for your time.

Start by noting that
$\displaystyle 4t^2 + \frac{1}{t^2} + 4$
is a perfect square.

Hint: It is of the form
$\displaystyle \left ( at + \frac{b}{t} \right )^2$

-Dan
• Jun 2nd 2008, 05:06 PM
topsquark
Quote:

Originally Posted by Mathstud28
$\displaystyle \int(2t)^2+t^{-2}+4dt$

simplifying to

$\displaystyle \int{4t^2+t^{-2}+4dt}=\frac{4}{3}t^3-t^{-1}+4t+C$

You left off the square root.

-Dan
• Jun 2nd 2008, 05:06 PM
Krizalid
Not big deal, since $\displaystyle \sqrt{4t^{2}+\frac{1}{t^{2}}+4}=\frac{\sqrt{4t^{4} +4t^{2}+1}}{t}=\frac{2t^{2}+1}{t}.$

• Jun 2nd 2008, 05:07 PM
Mathstud28
Quote:

Originally Posted by topsquark
You left off the square root.

-Dan

where is there a...oh...sorry...its hard to see those little things
• Jun 2nd 2008, 05:07 PM
Ihatemccormick
first of all, the 1st term inside the root is it (2t)^2 or just 2t^2, if there is a bracket round 2t then you have to make it to 4t^2

edit: wait, my mistake I'm just being stupid lol
• Jun 2nd 2008, 05:37 PM
zellster87
Quote:

Originally Posted by topsquark
Start by noting that
$\displaystyle 4t^2 + \frac{1}{t^2} + 4$
is a perfect square.

Hint: It is of the form
$\displaystyle \left ( at + \frac{b}{t} \right )^2$

-Dan

Thanks for the quick replies guy, I really appreciate it.

Now when I take the integral of (2t^2 + 1) / t from 1 to e

I simplified it to ∫ 2t^2 / t + 1/t dt -----> ∫2t + t^-1 dt

But I have a problem with the "t^-1" term when I take the anti-derivative of it. It comes out as t^0 / 0 which cant be right. Did i simplify this wrong?

Thanks
• Jun 2nd 2008, 05:40 PM
topsquark
Quote:

Originally Posted by zellster87
Thanks for the quick replies guy, I really appreciate it.

Now when I take the integral of (2t^2 + 1) / t from 1 to e

I simplified it to ∫ 2t^2 / t + 1/t dt -----> ∫2t + t^-1 dt

But I have a problem with the "t^-1" term when I take the anti-derivative of it. It comes out as t^0 / 0 which cant be right. Did i simplify this wrong?

Thanks

$\displaystyle \int \frac{dt}{t} = ln|t| + C$

-Dan
• Jun 2nd 2008, 05:42 PM
zellster87
Woah, I really am rusty. Thanks for you help guys.