Question

evaluate the following: integral S (e^3x)/((e^6x)+1) dx
(hint: u=e^3x)

Answer 1

\[u=e^{3x} \implies du=3e^{3x}dx\]
\[I=\frac{1}{3}\int \frac{du}{u^2+1}=\frac{1}{3}\tan^{-1}u+c=\frac{1}{3}\tan^{-1}(e^{3x})+c.\]

Answer 2

\[\int\limits_{}^{}\frac{e^{3x}}{e^{6x}+1} dx\]

Thinking....
\[e^{3x}=\tan(\theta)\]
=>
dang it

Answer 3

great job math lol

Answer 4

Obviously \(I\) is the integral you gave.

Answer 5

@myin: My substitution was a hint from the question :-D

Answer 6

\[\int\limits_{}^{}\frac{ \frac{1}{3} \sec^2(\theta)}{\tan^2(\theta)+1} d \theta\]

Answer 7

i like my sub better

Answer 8

it contains both subs

Answer 9

haha that's awesome. i knew it was (1/3)arctan(e3x) + C. i was just seeing if this actually worked :)

Answer 10

\[\int\limits \frac{e^{3 x}}{1+e^{6 x}} \, dx=\frac{1}{3} \text{ArcTan}\left[e^{3 x}\right]+c \]