# Math Help - Definite integral ln(1+t)dt

1. ## 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..

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

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

This led me to the following

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

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 $U=ln(1+t)$ and $U=1/(1+t)$ incorrect?

2. 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..

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

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

This led me to the following

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

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 $U=ln(1+t)$ and $U=1/(1+t)$ incorrect?
You're correct...but you need to think a little "outside of the box" for this integral.

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

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

Thus, $\,du=\,dt$

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

Can you take it from here?

--Chris

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

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

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

Thus, $\,du=\,dt$

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

Can you take it from here?

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

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

--Chris

4. 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).

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

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

$\int_1^6 \ln u~du$

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

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)

5. 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)