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

The integral from-1 to 1 of (e^x)/(e^x+1) is 1

Popular >

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

\int_{- 1}^{1} \frac{e ^{x}}{e ^{x} + 1} d x

1

Solution steps

\int_{- 1}^{1} \frac{e ^{x}}{e ^{x} + 1} d x

Apply u-substitution

: \int_{\frac{1}{e} + 1}^{e + 1} \frac{1}{u} d u

= \int_{\frac{1}{e} + 1}^{e + 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{\frac{1}{e} + 1}^{e + 1}

Compute the boundaries:

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

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

Simplify

= 1

\int_{1}^{3} \frac{x ^{2} + x - 1}{x ^{2}} d x

\int_{0}^{1} x \cdot arctan (x) d x

\int_{- 2}^{7} \frac{1}{( x + 1 ) ^{\frac{2}{3}}} d x

\int_{4}^{8} \frac{x}{x ^{2} - 15} d x

\int_{0}^{1} e^{x^{2}} (x^{3} + x) d x

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