1. ## integrate (ln(x+t)) dt

integrate (ln(x+t)) dt - Wolfram|Alpha

how did it move from the penultimate step to the last?

2. ## Re: integrate (ln(x+t)) dt

Originally Posted by BabyMilo
integrate (ln(x+t)) dt - Wolfram|Alpha

how did it move from the penultimate step to the last?
Before answering that, is x a function of t? Or is it being treated as constant?

3. ## Re: integrate (ln(x+t)) dt

I am solving for this.

4. ## Re: integrate (ln(x+t)) dt

Originally Posted by Prove It
Before answering that, is x a function of t? Or is it being treated as constant?
What difference does it make :S

5. ## Re: integrate (ln(x+t)) dt

Originally Posted by BabyMilo
integrate (ln(x+t)) dt - Wolfram|Alpha

how did it move from the penultimate step to the last?
Assuming that x is a constant value :

first : substitute $x+t=p$ , so $dt=dp$

Hence :

$I=\int \ln p \,dp$

Now use integration by parts to solve this integral . Hint :

$u=\ln p ~\text{ and }~ dv=dp$

6. ## Re: integrate (ln(x+t)) dt

Originally Posted by princeps
Assuming that x is a constant value :

first : substitute $x+t=p$ , so $dt=dp$

Hence :

$I=\int \ln p \,dp$

Now use integration by parts to solve this integral . Hint :

$u=\ln p ~\text{ and }~ dv=dp$
you wrote what wolfram wrote in different symbols, didnt you?

anyway, my question is how did it move from the penultimate step to the last?

to

Many thanks.

7. ## Re: integrate (ln(x+t)) dt

Originally Posted by BabyMilo
you wrote what wolfram wrote in different symbols, didnt you?

anyway, my question is how did it move from the penultimate step to the last?

to

Many thanks.
Actually solution is :

$I=(t+x)(\ln(t+x)-1)+C$

Hence :

$I=(t+x)\cdot \ln (t+x)-t+(C-x)$

Since x is a constant value we have that :

$I=(t+x)\cdot \ln (t+x)-t+C'$