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

2 Answers
Jan 7, 2017

Answer:

#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#

Jan 7, 2017

Answer:

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#