Integrating ln(t+1) from 0 to e^2x?

[Enujybab]

Homework Statement

How would I begin to integrate ln(t+1) from 0 to e^2x?

Homework Equations

d/dx[log base a of u]=1/(lna)u du/dx

Can the original equation be manipulated to use this derivative?

The Attempt at a Solution

Not sure where to start.

 

[Dick]

It's the sort of problem where you could actually guess the antiderivative, but if you can't, integrate by parts.

 

[masterslave]

It's an integration by part question. First, use a substitution to get it to one variable instead of a polynomial in the logarithm.
w=t+1
dw=dt
Substitute (I'm going to revert the limits in the end to the original variable, so you know):
Integrate of ln(w)dw
Let:
u=ln(w) and dv=dw
du=(1/w)*dw and v=w

w*ln(w)-Int(w*(1/w)*dw)
w*ln(w)-Int(dw)
w*ln(w)-w
(t+1)*[ln(t+1)-1]
now evaluate at your endpoints:
{(e^2x+1)*[ln(e^2x+1)-1]}+{1}
You can probably simplify this some more, but there it is.