please i really cant do it. i think you have you use integration by substitution.
I assume you meant ∫ e^(x) dx / [1 + e^(x)]
Let u = 1 + e^(x), du = e^(x) dx
∫ e^(x) dx / [1 + e^(x)]
= ∫ du / u
= ln(u) + C
= ln(1 + e^(x)) + C
Division always comes before addition unless there are brackets. Integration by substitution is not required for this integral.
â« (eË / 1 + eË) dx
â« (eË + eË) dx
â« 2eË dx
2 â« eË dx
2eË + C
You can rewrite it as e^x(1+e^x)^-1dx
Then you can set u = 1 + e^x, giving you:
du = e^xdx
So you have the integral of (u^-1)du, which is ln(u), which is ln(1 + e^x) + C.
The answer is ln(1 + e^x) + C.
Comments
I assume you meant ∫ e^(x) dx / [1 + e^(x)]
Let u = 1 + e^(x), du = e^(x) dx
∫ e^(x) dx / [1 + e^(x)]
= ∫ du / u
= ln(u) + C
= ln(1 + e^(x)) + C
Division always comes before addition unless there are brackets. Integration by substitution is not required for this integral.
â« (eË / 1 + eË) dx
â« (eË + eË) dx
â« 2eË dx
2 â« eË dx
2eË + C
You can rewrite it as e^x(1+e^x)^-1dx
Then you can set u = 1 + e^x, giving you:
du = e^xdx
So you have the integral of (u^-1)du, which is ln(u), which is ln(1 + e^x) + C.
The answer is ln(1 + e^x) + C.