Integrating an ln() Function.

**Acidicism** · May 4th 2009, 02:29 PM

Hey, I need a help integrating an natural log function, as the title states. My teacher never taught it to the class and I am having a difficult time trying to understand it on other sites. My AB test is in 2 days and I am very worried about getting a low score. The question: v(t) = ln(1 + t^2) How far does the object travel between time t = 0 and time t = 10? When I put the integration of the function with limits 0 and 10, the calculator does not spit out the one number that I want. I have the TI-89 and when I put the MODE -> APPROX or when I (diamond) [ENTER] the equation, it still does not spit out one number. I would greatly appreciate help for both of my concerns

**skeeter** · May 4th 2009, 02:42 PM

Originally Posted by Acidicism

Hey, I need a help integrating an natural log function, as the title states. My teacher never taught it to the class and I am having a difficult time trying to understand it on other sites. My AB test is in 2 days and I am very worried about getting a low score. The question: v(t) = ln(1 + t^2) How far does the object travel between time t = 0 and time t = 10? When I put the integration of the function with limits 0 and 10, the calculator does not spit out the one number that I want. I have the TI-89 and when I put the MODE -> APPROX or when I (diamond) [ENTER] the equation, it still does not spit out one number. I would greatly appreciate help for both of my concerns

put it back to AUTO mode ... then do "diamond" ENTER

my TI-89 says $\int_0^{10} \ln(1+t^2) \, dt = 29.093$

**Soroban** · May 4th 2009, 03:02 PM

Hello, Acidicism!

To integrate a function with $\ln$ , "by parts" is usually needed.

$v(t) \:=\: \ln(t^2 + 1)$
How far does the object travel between time $t = 0$ and $t = 10$ ?

By parts: . $\begin{array}{ccccccc}u &=& \ln(t^2+1) & & dv & = & dt \\ \\[-4mm] du &=& \dfrac{2t\,dt}{t^2+1} & & v &=& t \end{array}$

Then: . $\int\ln(t^2+1)\,dt \;=\;t\ln(t^2+1) \;- \int\frac{2t^2}{t^2+1}\,dt \;=\;t\ln(t^2+1) \;- \int\left[2 - \frac{2}{t^2+1}\right]\,dt$

. . $= \;t\ln(t^2+1) - 2t + 2\arctan(t)\,\bigg]^{10}_0$

. . $= \;\bigg[10\ln(101) - 2(10) + 2\arctan(10)\bigg] - \bigg[0\ln(1) - 2(0) + 2\arctan(0)\bigg]$

. . $= \;10\ln(101) -20 + 2\arctan(10)$

. . $= \;20.09346052...$

**Acidicism** · May 4th 2009, 03:03 PM

Oh, I just realized I integrated it with respect to x instead of t. Thanks sir.

I don't really get this; why is there a t in front and why are you subtracting the integral of the stuff inside?

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