Question

Integral (e^x + e^-x)^2 dx

Solve the integral, What should I choose for u subsitution?

Answer 1

multiply it out

Answer 2

okay let me check it out.. thanks

Answer 3

Yea im still stuck after that..
I'm getting (That 2nd term is supposed to be (e^(-x)) \[\int\limits_{}^{}(e^x)^2+2(e^x)(e^-x))+(e^-x))^2\]

Answer 4

\[(e^x)^2=e^{2x}\]

Answer 5

note that:
\[e^{x}e^{-x} = e^{x-x} = e^0 = 1\]

Answer 6

can you help me a little more, So I understand the algebra to make that equation but what u would i use to substitute

Answer 7

1/2e^2x+2-1/2e^-2x+c

Answer 8

dont sub. just evaluate

Answer 9

you are missing a 2x

Answer 10

I have: \[\int\limits1+2e^xe^-x dx\]

Answer 11

e^{-x}
\[e^{-x}\]
;)

Answer 12

plus 2x you are correct z

Answer 13

After you multiply it out, you want to break up the integral to see it easier:
\[\int\limits_{}^{}e^{2x}+2+e^{-2x}dx = \int\limits e^{2x}dx+\int\limits2dx+\int\limits e^{-2x}dx\]

Answer 14

Ahhh, I'm going to do a u subsitution for 2x and then make du = 2dx then evaluate \[\int\limits(e^u+e^{-u})du\]

Answer 15

looks like you are missing some stuff there.

Answer 16

hmm, well I have \[\int\limits(e^{2x}+2+e^{-2x})dx\]
Then I set u = 2x , solve for du =2dx, and rewrite the integral, as i had above. Does that work?

Answer 17

\[\int\limits(e^u+2+e^{-u})\frac{du}{2}\]

Answer 18

Where did the 1/2 come from?

Answer 19

\[du=2dx\]

so

\[dx=\frac{du}{2}\]

Answer 20

Hmmm Im pretty lost right now.. Bnut got it correct up there (he was missing a 2x though ) Can you walk me through how to get there?

Answer 21

\[\int\limits(e^{2x}+2+e^{-2x})dx\]

let \[u=2x\]

\[du=2dx\Rightarrow dx=\frac{du}{2}\]


\[\int\limits(e^{u}+2+e^{-u})\frac{du}{2}\]

\[=\frac{1}{2}\int\limits(e^{u}+2+e^{-u})du\]

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

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

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

Answer 22

Bravo! Thanks

Answer 23

np..night ;)