Question

Evaluate the indefinite integral:
(e^(2x) -1)/(e^(2x) + 1) dx
A hint was given to multiply top and bottom by e^(-x). And we're only supposed to use a "u" substitution.... as in, we can only use what we have learned in class so far which has only been up to u substitutions at this point.

Answer 1

going with that hint\[\frac{e^{2x}-1}{e^{2x}+1}\frac{e^{-x}}{e^{-x}}=\]?

Answer 2

\[\frac{e^{2x}-1}{e^{2x}+1}\frac{e^{-x}}{e^{-x}}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=\frac{\sinh x}{\cosh x}\]

Answer 3

let \[u=\cosh x\]

Answer 4

are u familiar with hyperbolic functions like sinh and cosh?

Answer 5

No, we haven't gotten to them yet.

Answer 6

ok\[\frac{e^{2x}-1}{e^{2x}+1}\frac{e^{-x}}{e^{-x}}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\]let \(u=e^{x}+e^{-x}\)

Answer 7

Try let u=2x du=2dx
\[\frac{1}{2}\int\limits \frac{e^u-1}{e^u+1} \, du\]
Separate the top \(e^u\) and -1 from numerator,
\[\frac{1}{2}\int\limits \left(\frac{e^u}{e^u+1}-\frac{1}{e^u+1}\right) \, du\]
Can you do it now?

Answer 8

sam what will be\[\int\frac{1}{1+e^u} \text{d}u\]?

Answer 9

\(\int\frac{e^{-u}}{1+e^{-u}} \text{d}u\)

Answer 10

Since you differentiate e^u is e^u, try another u-subbing then separate

Answer 11

seeing this is harder than the first one :)

Answer 12

Yeah but he/she didn't learn about hyp functions

Answer 13

We'll have to use a few u-subs here

Answer 14

yeah and i fixed that with u=e^x+e^-x :)

Answer 15

@mukushla Isn't \[\frac{e^x - e^-x }{2} = \sinh x\] ?

Answer 16

My bad. The 2 from sinh x and cosh x cancels.