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

Question

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

mr.math · Answer

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

myininaya · Answer

$$\int\limits_{}^{}\frac{e^{3x}}{e^{6x}+1} dx$$

Thinking....
$$e^{3x}=\tan(\theta)$$
=>
dang it

myininaya · Answer

great job math lol

mr.math · Answer

Obviously $I$ is the integral you gave.

mr.math · Answer

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

myininaya · Answer

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

myininaya · Answer

i like my sub better

myininaya · Answer

it contains both subs

anonymous · Answer

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

anonymous · Answer

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