Question

Integrate (e^2x)/(1 + e^x) @Calculus1

Answer 1

u sub it via
\[u=e^x, du = e^xdx\] to get
\[\int \frac{u}{u+1}du\]

Answer 2

what happened to the two sat?

Answer 3

oh
\[e^x\times e^xdx=e^{2x}\] in the numerator

Answer 4

sorry i meant
\[e^{2x}dx\]

Answer 5

that is why it is just
\[\int\frac{udu}{u+1}\]

Answer 6

still have to integrate though. since power is the same top and bottom divide to get
\[\frac{u}{u+1}=1-\frac{1}{u+1}\] then integrate to get
\[u-\ln(u+1)\] and finally
\[e^x-\ln(e^x+1)\]

Answer 7

+C
!!!!!!!!

Answer 8

I'm still not understanding why the 2 is irrelevant...

Answer 9

He integrated the form above, so it is relevant. If it was irrelevant you would integrate without a u in the denominator, just a du. because he wrote \[u=e^x \to du=e^xdx\]the numerator is now \[udu=e^{2x}dx\]Then he did his fraction trick to the expression and was able to integrate it.

Answer 10

Can you show a bit more of how the fraction was split?

Answer 11

I was just wondering and figure it out... clever\[{u \over u+1}={u+1-1\over u+1}={u+1\over u+1}-{1\over u+1}=1-{1\over u+1}\]That's why satellite gets props from me...

Answer 12

oh thank you for the compliment. but in fact i did not do that trick. i just divided

Answer 13

how did you "just divide" without a 1 in the numerator? I've never done that.

Answer 14

i will see if i can write it it is one step

Answer 15

ok i cant write division here but it does like this basically

you divide u by u + 1
u goes in to u one time
1 times u + 1 is u + 1
subtract and get -1

that is the remainder. so answer is
\[1-\frac{1}{u+1}\] i can never figure out how to write a long division here

Answer 16

Thanks for the lessons sat and turing!

Answer 17

Oh I see, that makes sense, I'm surprised I've never seen it before. I guess you could draw itThanks!!!!!!! :D