# Definite integral ln(1+t)dt

• Sep 25th 2008, 06:04 PM
mkelly09
Definite integral ln(1+t)dt
I need some help with a particular question that i can't seem to figure out no matter how much time i've spent on it this evening..

$\displaystyle \int_{0}^{5} ln(1+t)dt$

My first way of thinking led me to a dead end. I used integration by parts with $\displaystyle U=ln(1+t)$ and $\displaystyle U=1/(1+t)$ and $\displaystyle V=1dt$ and $\displaystyle V=t$

This led me to the following

$\displaystyle = [tln(1+t)]_{0}^{5} - \int_{0}^{5} t/(1+t)dt$

and this is where i'm stuck.. can't seem to figure out the integration for the second integral here.

I think my line of thinking is wrong.. is my assumption that $\displaystyle U=ln(1+t)$ and $\displaystyle U=1/(1+t)$ incorrect?
• Sep 25th 2008, 06:51 PM
Chris L T521
Quote:

Originally Posted by mkelly09
I need some help with a particular question that i can't seem to figure out no matter how much time i've spent on it this evening..

$\displaystyle \int_{0}^{5} ln(1+t)dt$

My first way of thinking led me to a dead end. I used integration by parts with $\displaystyle U=ln(1+t)$ and $\displaystyle U=1/(1+t)$ and $\displaystyle V=1dt$ and $\displaystyle V=t$

This led me to the following

$\displaystyle = [tln(1+t)]_{0}^{5} - \int_{0}^{5} t/(1+t)dt$

and this is where i'm stuck.. can't seem to figure out the integration for the second integral here.

I think my line of thinking is wrong.. is my assumption that $\displaystyle U=ln(1+t)$ and $\displaystyle U=1/(1+t)$ incorrect?

You're correct...but you need to think a little "outside of the box" for this integral.

$\displaystyle \int\frac{t}{1+t}\,dt$

Let $\displaystyle u=t+1\implies t=u-1$

Thus, $\displaystyle \,du=\,dt$

Therefore, the integral becomes $\displaystyle \int\frac{t\,du}{u}=\int\frac{u-1}{u}\,du=\int\left(1-\frac{1}{u}\right)\,du$

Can you take it from here?

--Chris
• Sep 25th 2008, 06:54 PM
Chris L T521
Quote:

Originally Posted by Chris L T521
You're correct...but you need to think a little "outside of the box" for this integral.

$\displaystyle \int\frac{t}{1+t}\,dt$

Let $\displaystyle u=t+1\implies t=u-1$

Thus, $\displaystyle \,du=\,dt$

Therefore, the integral becomes $\displaystyle \int\frac{t\,du}{u}=\int\frac{u-1}{u}\,du=\int\left(1-\frac{1}{u}\right)\,du$

Can you take it from here?

--Chris

Here's another way I just thought of: add "zero" to the numerator

$\displaystyle \int\frac{t}{t+1}\,dt=\int\frac{t+1-1}{t+1}\,dt=\int\left[\frac{t+1}{t+1}-\frac{1}{t+1}\right]\,dt=\int\left(1-\frac{1}{t+1}\right)\,dt$

--Chris
• Sep 25th 2008, 07:05 PM
Jhevon
using integration by parts on that guy directly is overkill. (that is, if you know the integral of ln(x) by heart--which you should).

$\displaystyle \int_0^5 \ln (1 + t)~dt$

Let $\displaystyle u = t + 1$, this substitution yields

$\displaystyle \int_1^6 \ln u~du$

which we know is $\displaystyle u \ln u - u \bigg|_1^6$ ...

see post #7 here for how to integrate ln(x) using integration by parts. note that you could also figure out the integral by contemplating the derivative of xln(x)
• Sep 26th 2008, 06:02 AM
mkelly09
Thanks for your replies, they certainly solved my problem. In fact, now i know of a few different ways of doing it when before i couldn't figure out one.

I think i will be committing to memory the antiderivative of ln(x)