Question

If `d/dx(F(x))=1/(e^x+1)`, then find F(x) given that F(0) = log `1/2`.

Answer 1

Given

`d/dx(F(x))=1/(e^x+1)`

F(x) = `int1/(e^x+1)dx`

Let e^x = t

e^x dx = dt

dx = `dt/e^x`

= `dt/t`

F(x) = `int1/(t(t+1))dt`

`1/(t(t+1))=A/t+B/(t+1)`

I = A (t +1) +Bt

I = At + A + Bt

A + B = 0

A = 1

B = –1

F(x) = `int1/(t(t+1))dt`

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

= log|t| – log|t + 1| + C

= `log|t/(t+1)|+C`

F(x) = `log|e^x/(e^x+1)|+C`

At F(0) = log `1/2` ; x = 0

F(0) = `log|e^0/(e^0+1)|+C`

= `log  1/2+C`

⇒ `log  1/2=log1/2+C`

⇒ C = 0

F(x) = `log|e^x/(e^x+1)|`