Exponential integration

[Ian McPherson]

My professor gave us a practice problem e^3x
____ dx
e^3x + 1

The answer given is 1/3 ln (e^3x + 1) +c

would u = 3x? or would U = ^3x + 1. I haven't seen one like this before. Can you please explain how he reached this answer?

 

[soroban]

Hello, Ian McPherson!

\(\displaystyle \displaystyle\text{My professor gave us a practice problem: }\:\int\frac{e^{3x}\,dx}{e^{3x} + 1}\)

\(\displaystyle \text{The answer given is: }\:\frac{1}{3}\ln\left(e^{3x}+ 1\right) + C\)

\(\displaystyle \text{Let }\,u \:=\:e^{3x}+1 \quad\Rightarrow\quad du \:=\:3e^{3x}dx \quad\Rightarrow\quad e^{3x}dx \:=\:\frac{1}{3}du\)

\(\displaystyle \displaystyle \text{Substitute: }\:\int\frac{\frac{1}{3}du}{u} \;=\;\tfrac{1}{3} \int\frac{du}{u} \;=\;\tfrac{1}{3}\ln(u) + C\)

\(\displaystyle \text{Back-substitute: }\:\frac{1}{3}\ln\left(e^{3x}+1\right) + C\)

 

[Ian McPherson]

That's a bit confusing looking! But i think i get it. Thank you.