What is the integral from 0 to 2 of 1/(1+e^{2x)} ?

The integral from 0 to 2 of 1/(1+e^{2x)} is -1/2 (ln(1+e^4)-ln(2))+2

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integral from 0 to 2 of 1/(1+e^{2x)}

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

- \frac{1}{2} (ln (1 + e^{4}) - ln (2)) + 2

Decimal Notation

0.33749 \dots

Solution steps

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

\frac{1}{1 + e ^{2 x}} = - \frac{e ^{2 x}}{1 + e ^{2 x}} + 1

= \int_{0}^{2} - \frac{e ^{2 x}}{1 + e ^{2 x}} + 1 d x

Apply the Sum Rule:

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

= - \int_{0}^{2} \frac{e ^{2 x}}{1 + e ^{2 x}} d x + \int_{0}^{2} 1 d x

\int_{0}^{2} \frac{e ^{2 x}}{1 + e ^{2 x}} d x = \frac{1}{2} (ln (1 + e^{4}) - ln (2))

\int_{0}^{2} 1 d x = 2

= - \frac{1}{2} (ln (1 + e^{4}) - ln (2)) + 2

x \to - \infty lim (- 5 x^{3})

\int \frac{3 x ^{3} + 4 x}{( x ^{2} + 5 ) ^{2}} d x

\int \frac{3 x ^{4} - 2 x + 5}{x + 1} d x

\frac{d}{d x} (A x)

\frac{d}{d y} (1 - x y cos (π y))

What is the integral from 0 to 2 of 1/(1+e^{2x)} ?
The integral from 0 to 2 of 1/(1+e^{2x)} is -1/2 (ln(1+e^4)-ln(2))+2