What is the antiderivative of 7/7+e^x?

Question

Can someone please tell me what the antiderivative of 7/(7+e^x) is. Thank you

Answer 1

7 / (7+e^x)
= 1 - e^x/(7+e^x)

Substituting u = 7+e^x, du = e^x * dx we get for the second term:
(e^x / (7 + e^x)) dx = du/u
which integrates to ln|u| = ln|7+e^x|. So,

int(1 - e^x/(7+e^x)) dx = int(1)dx - int(e^x/(7+e^x)) dx
= x - ln|7+e^x| + C.

Answer 2

Integral ( 7/(7 + e^x) ) dx

Let u = e^x + 7
u - 7 = e^x

ln(u - 7) = x, therefore
[1/(u - 7)]du = dx

So we have

Integral ( (7/u) [1/(u - 7)] du )

Let's pull out the constant 7. This will give us

7 * Integral ( (1/u) [1/(u - 7)] du)

We can merge those fractions as 1.

7 * Integral ( 1/[u(u - 7)] ) du

At this point, you need to complete the square in the denominator upon multiplying. I'm going to stop here because it gets too complicated.

Answer 3

integral 7/7 +e^x
= integral 1 +e^x
= x +e^x +C

Sorry, just saw you corrected this to 7/(7+e^x)
integral(7/(7+e^x) dx
= 7*integral(dx/(7+e^x))
Let u 7+e^x, then dx = du, so we have:
7*integral(du/u) = 7* ln u + C = 7*ln (7+e^x) + C

Answer 4

benoit3535 beat me to it, he's right with x-ln(7+e^x)+c

Answer 5

=x+e^x+c