# What is the function y if dy/dx=1/(e^x+1)?

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### 2 Answers

It is given that dy/dx = 1/(e^x + 1)

y = Int[ 1/(e^x + 1) dx]

=> Int[(e^x + 1 - e^x)/(e^x + 1) dx]

=> Int[(e^x +1)/(e^x+1) - e^x/(e^x + 1) dx]

=> Int[(e^x +1)/(e^x+1) dx] - Int[e^x/(e^x + 1) dx]

=> Int[1 dx] - Int[e^x/(e^x + 1) dx]

Int [ 1 dx ] = x

Int[e^x/(e^x + 1) dx]

let e^x + 1 = t

dt = e^x dx

=> Int [ 1/t dt]

=> ln t

substitute t = e^x + 1

=> ln |e^x + 1|

**The final integral is x - ln|e^x +1| + C**

To find the function y, we'll have to determine the indefinite integral of the given function dy/dx.

I'll suggest to replace e^x by t.

e^x = t => x = ln t

We'll differentiate both sides:

dx = dt/t

We'll re-write the integral in t:

Int dx/(e^x+1) = Int dt/t*(t+1)

We'll decompose the fraction 1/t(t+1) in a difference of partial fractions.

1/t(t+1) = 1/t - 1/(t+1)

Int dt/t*(t+1) = Int dt/t - Int dt/(t+1)

Int dt/t*(t+1) = ln |t| - ln|t+1| + C

We'll apply quotient rule of logarithms:

Int dt/t*(t+1) = ln |t/(t+1)| + C

**The primitive function is: y = Int dx/(e^x+1) = ln e^x/(e^x+1) + C.**