[SOLVED] Calculus: Integration by substitution

**juicysharpie** · February 4th 2009, 09:12 AM

I'm stuck...Could someone please help explain how to do this problem to me?

Find by substitution: the integral e to e^2 of dx/x(lnx)^2

**Plato** · February 4th 2009, 09:18 AM

$\int_e^{e^2 } {\frac{{dx}}<br /> {{x\left( {\ln (x)} \right)^2 }}} = \left. {\frac{{ - 1}}<br /> {{\ln (x)}}} \right|_e^{e^2 }$

**Moo** · February 4th 2009, 09:18 AM

Hello,

Originally Posted by juicysharpie

I'm stuck...Could someone please help explain how to do this problem to me?

Find by substitution: the integral e to e^2 of dx/x(lnx)^2

Just substitute $t=\ln(x)$

**juicysharpie** · February 4th 2009, 09:39 AM

I'm sorry. I still don't understand...

**Krizalid** · February 4th 2009, 09:44 AM

It's basic stuff according to substitution method. What is exactly you don't get?

**juicysharpie** · February 4th 2009, 09:49 AM

I found u=ln x. du/dx = 1/x and du= 1/x dx. I found the new limits, 1 (lower) 2 (upper). So the new equation is the integral from 1 to 2 of 1/u^2 du. How do I derivate that so that I can plug in 1 and 2 to solve? I tried ln u x ln u but I don't think that's correct because I did not get 1/2 like the answer is supposed to be...

**Plato** · February 4th 2009, 10:00 AM

Originally Posted by juicysharpie

I found u=ln x. du/dx = 1/x and du= 1/x dx. I found the new limits, 1 (lower) 2 (upper). So the new equation is the integral from 1 to 2 of 1/u^2 du.

$\int_1^2 {\frac{{du}}<br /> {{u^2 }}} = \int_1^2 {u^{ - 2} du} <br />$

**juicysharpie** · February 4th 2009, 10:11 AM

oh, i see...thanks for the help!!

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