Question

can someone show me how to calculate the integral

Answer 1

\[\int\limits (e^7x)/(e^7x-9e^-7x) dx\]

Answer 2

its 9e^-7x

Answer 3

You have to put curly braces around the exponent stuff if it's more than one character.
Like this e^{-7x}.

\[\large\rm \int\limits\frac{e^{7x}}{e^{7x}-9e^{-7x}}dx\]This is the integral we're dealing with? Hmm.

Answer 4

yes

Answer 5

What happens if we make a u-sub,\[\large\rm u=e^{7x}-9e^{-7x}\]I think this should lead somewhere nice.
Do you think you can come up with the du?

Answer 6

Ohhh I'm mistaken, that doesn't work so well does it >.<
Grr rethinking..

Answer 7

\[du=7e^7x+73e^{-7x} dx\]

Answer 8

63 *

Answer 9

Maybe we can do this,\[\large\rm \int\limits\limits\frac{e^{7x}}{e^{7x}-9\left(e^{7x}\right)^{-1}}dx\]And make the substitution,\[\large\rm u=e^{7x}\]

Answer 10

do you think simplifying denominator works ? but not sure how

Answer 11

Do you understand the exponent rule I applied?
That should help us out a lot.

Answer 12

\[\frac{ 1 }{ 7 } \int\limits \frac{ 1 }{ u-9u^{-1}} du\]

Answer 13

is what i came up after

Answer 14

Mmm ok great!
Multiplying through by u/u gives us\[\large\rm \frac17\int\limits\frac{u}{u^2-9}du\]

Answer 15

Hopefully you understood that last step.
Any ideas from this point?

Answer 16

thank you very much i figured it out

Answer 17

Cool :)
And welcome to OpenStudy!

Answer 18

Zedpdrix can you show me the last step how to get the final answer .. the answer i figured out is wrong last night but the system does not tell me the final answer thank you

Answer 19

From where we left off,\[\large\rm \frac17\int\limits\limits\frac{u~du}{u^2-9}\]
applying another substitution,\[\large\rm m=u^2-9\]\[\large\rm \frac12dm=u~du\]gives us,\[\large\rm \frac{1}{7}\int\limits\frac{\frac12dm}{m}\]Combining the fractions and integrating,\[\large\rm \frac{1}{14}\ln|m|+c\]Undoing our first substitution,\[\large\rm \frac{1}{14}\ln|u^2-9|+c\]Undoing our other substitution,\[\large\rm \frac{1}{14}\ln\left|\left(e^{7x}\right)^2-9\right|+c\]Applying exponent rule,\[\large\rm \frac{1}{14}\ln\left|e^{14x}-9\right|+c\]

Answer 20

For some reason Wolfram is showing the solution as\[\large\rm \frac{1}{14}\ln\left(9-e^{14x}\right)+c\]Reversing the terms in the log,
and dropping the absolutes... hmm

Answer 21

I'm not sure if that matters though.