What is the integral from-1 to 0 of 1/(1+e^t) ?

The integral from-1 to 0 of 1/(1+e^t) is -ln(2)+ln(1+1/e)+1

Popular >

integral from-1 to 0 of 1/(1+e^t)

\int_{- 1}^{0} \frac{1}{1 + e ^{t}} d t

- ln (2) + ln (1 + \frac{1}{e}) + 1

Decimal Notation

0.62011 \dots

Solution steps

\int_{- 1}^{0} \frac{1}{1 + e ^{t}} d t

\frac{1}{1 + e ^{t}} = - \frac{e ^{t}}{1 + e ^{t}} + 1

= \int_{- 1}^{0} - \frac{e ^{t}}{1 + e ^{t}} + 1 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 1}^{0} \frac{e ^{t}}{1 + e ^{t}} d t + \int_{- 1}^{0} 1 d t

\int_{- 1}^{0} \frac{e ^{t}}{1 + e ^{t}} d t = ln (2) - ln (1 + \frac{1}{e})

\int_{- 1}^{0} 1 d t = 1

= - (ln (2) - ln (1 + \frac{1}{e})) + 1

Simplify

= - ln (2) + ln (1 + \frac{1}{e}) + 1

\int_{0}^{\frac{1}{2}} x^{2} e^{6 x} d x

\int_{- 2}^{2} 28 - 7 x^{2} d x

\int_{- 1}^{1} (x^{4} - 2 x^{2} + 1) d x

\int_{0}^{4} (5 x + 3) d x

\int_{0}^{1} (x + 1) sin (nπ x) d x

What is the integral from-1 to 0 of 1/(1+e^t) ?
The integral from-1 to 0 of 1/(1+e^t) is -ln(2)+ln(1+1/e)+1