# Thread: integrals involving multiple e's

1. ## integrals involving multiple e's

The problem is: $\int \frac{e^{3x}}{e^{6x}+1}$

By using multiple product rules, I ended up with $\frac{5^{2x}}{x}+2ln(5)ln(3x)$

The answer choice I selected had the same answer, except that it had $5^{2x}$ at the end. $\frac{5^{2x}}{x}+2ln(5)ln(3x)5^{2x}$

Which one is correct? If it is the second one, can you please explain where the additional $5^{2x}$ came from?

2. Originally Posted by TheMathTham
The problem is: $\int \frac{e^{3x}}{e^{6x}+1}$

By using multiple product rules, I ended up with $\frac{5^{2x}}{x}+2ln(5)ln(3x)$

The answer choice I selected had the same answer, except that it had $5^{2x}$ at the end. $\frac{5^{2x}}{x}+2ln(5)ln(3x)5^{2x}$

Which one is correct? If it is the second one, can you please explain where the additional $5^{2x}$ came from?
Use $u=e^{3x}$ and then remember the arctan

3. e^(6x) = [e^(3x)]^2

Let u = e^(3x)

you should obtain 1/3 arctan(e^3x) + C

what multiple product rules do you use on an integral?

4. oh wow i just realized i wrote the problem for one question and then the work to a different question that i was having trouble with

hehe ok so fail there by me...

The problem to the work was $\frac{\mathrm{d} }{\mathrm{d} x}(ln(3x))5^{2x})$. If you could help on this one too that would be great

And as for the original problem, I understand the answer now. But I'm curious, why is the $\int \frac{u}{u^{2}+1}=\arctan (u)$ ?

Sorry for the mess up

5. The second question first

for integral [ e^(3x)/(e^(6x)+1)dx]

if u = e^(3x) du = 3 e^(3x) dx

so you get 1/3 integral [1/(u^2 +1)du] which is the arctan(u)

For the first

[a^(u)] ' = ln(a) a^(u) du/dx

[5^(2x)] ' = ln(5)5^(2x)2

6. Originally Posted by Calculus26
The second question first

for integral [ e^(3x)/(e^(6x)+1)dx]

if u = e^(3x) du = 3 e^(3x) dx

so you get 1/3 integral [1/(u^2 +1)du] which is the arctan(u)

For the first

[a^(u)] ' = ln(a) a^(u) du/dx

[5^(2x)] ' = ln(5)5^(2x)2

For the second question, I was asking why the $\int \frac{u}{u^{2}+1}=\arccos (u)$. I thought that $\frac{u}{u^{2}+1}=\arccos (u)$ not the $\int \frac{u}{u^{2}+1}=\arccos (u)$

For the first question...I'm not following how you arrived there. What I did was to use the product rule and make $u=ln(3x), u{}'=\frac{1}{x}, v=5^{2x}, v{}'=2ln(5)$. I arrived at $v{}'$ by substituting $2xln(5)$ in for $5^{2x}$ and then using the product rule again in which the new variables were $u=2x, u{}'=2, v=ln(5), v{}'=0$. The final result was $\frac{5^{2x}}{x}+2ln(5)ln(3x)$. Is that correct?

7. for 5^(2x)

[5^(2x)] ' = ln(5)*5^(2x)*2

as in general [a^u]' = ln(a) *a^u du/dx

Eg (2^(x)) ' = ln(2) 2^x

(2^(sin(x)) ' = ln(2)2^(sin(x)) cos(x)

By the way integral ( u/(u^2+1)du) = 1/2 ln(u^2+1)

8. Originally Posted by Calculus26
for 5^(2x)

[5^(2x)] ' = ln(5)*5^(2x)*2

as in general [a^u]' = ln(a) *a^u du/dx

Eg (2^(x)) ' = ln(2) 2^x

(2^(sin(x)) ' = ln(2)2^(sin(x)) cos(x)

By the way integral ( u/(u^2+1)du) = 1/2 ln(u^2+1)
ohhhhhh i get it. thanks a ton!