How do you find the indefinite integral of int (3^(2x))/(1+3^(2x))?

2 Answers
Andrea S.
Jan 7, 2017

int 3^(2x)/(1+3^(2x))dx=1/(2ln3)ln abs(1+3^(2x))+C

Explanation:

Substitute t=3^(2x)=e^(2ln3x), dt= 2ln3*e^(2ln3x)

int 3^(2x)/(1+3^(2x))dx = 1/(2ln3)int (dt)/(1+t)= 1/(2ln3)ln abs(1+t)+C=1/(2ln3)ln abs(1+3^(2x))+C

Douglas K.
Jan 7, 2017

Do a u substitution. Please see the explanation.

Explanation:

int(3^(2x))/(1 + 3^(2x))dx =

int9^x/(1 + 9^x)dx

Let u = 1 + 9^x, then (du)/dx = (d(1))/dx + (d(9^x))/dx

The first term of the derivative is 0 but the second term requires logarithmic differentiation:

let y = 9^x, then (du)/dx = 0 + dy/dx

ln(y) = ln(9^x)

ln(y) = xln(9)

1/ydy/dx = ln(9)

dy/dx = ln(9)y

dy/dx = ln(9)9^x

Substituting this into the equation for (du)/dx

(du)/dx = ln(9)9^x

Writing this so that we can substitute into the integral:

9^xdx = 1/ln(9)du

Substituting this and u into the integral:

1/ln(9)int(du)/u = 1/ln(9)ln(u) + C

Reverse the substitution for u:

int(3^(2x))/(1 + 3^(2x))dx = 1/ln(9)ln(1 + 9^x) + C

Related questions
See all questions in Integrals of Exponential Functions
Impact of this question
7581 views around the world
You can reuse this answer
Creative Commons License