integrate e^2x / e^2x + 1 dx?

Answer 1

make u=e^2x + 1 then du=2e^2x dx and e^2x=u-1 then

integration of e^2x/(e^2x+1) becomes 1/2u du which is
ln(u)/2 + c and then you just insert the value of u into the equation and you get
(ln(e^2x+1))/2 + c

Answer 2

what is the first derivative of e^2x?
if f(x) = e^(2x) +1, then f'(x) = 2e^2x

So the above equation would be [f'(x) /2] / f(x)
the intergration of f'(x) / f(x) is ln f(x), but since the numirator of the derivative is divided by 2, then the integral is divided by 2,

so the intergration of e^2x / e^(2x) + 1 dx =

(1/2) ln | e^(2x) +1 |

This is only if you mean that e^(2x) +1, and not e^(2x+1)

Answer 3

?xe^(2x) dx from 0 to ? From here, we merely use an integration via section.... u = x du = dx dv = e^2x V = (e^2x) / 2 a million/2*[x*e^(2x)] - a million/2?e^2x dx from 0 to ? combine like conventional.... a million/2*[x*e^(2x)] - (a million/4)e^2x from -? to 0 Plug interior the limitations.... F(0) - F(-?) = [0 - (a million/4)] - [0] = -a million/4 very final answer: Converges to -a million/4

Answer 4

1+1 dx

Answer 5

Let e**2x +1 = y

dy/dx = 2 e**2x

so the term is int(dy/2ydx)dx = ln y/ 2 or in(e^2x+1)/2

Answer 6

take e^2x= t. then 2e^2x.(dx)=dt

in (integral)e^2x/e^2x+1 multiply and divide with two. then it will become
(1/2)(dt/(t+1)) which is equal to log((e^2x+1)^1/2)+ c

Answer 7

e^2x/e^2x+1=e^2x+1-1/e^2x+1
=1 - 1/e^2x+1
=1 - e^(-2x-1)
On integrating,
x + [e^(-2x-1)]/2 +c

Answer 8

Let u=e^(2x)+1.

Homework!

Answer 9

2x
since e^2x / e^2x =1

Answer 10

Glad maths were not like that when I was at school.