Answered: "Evaluate the integral. ∫01ex1+e2xdx"

opatovaL

Answered question

2021-10-19

Evaluate the integral.

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

Cristiano Sears

Skilled2021-10-20Added 96 answers

Step 1: Given that

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

We need to evaluate the integral
Step 2: Solving the integral
We have,

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

Substitute.

e^{x} = t

e^{x} \cdot d x = d t

\frac{d x}{d t} = e^{- x}

Plugging all the values into the integral we obtain,

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

= {[\tan^{- 1} (e^{x})]}_{0}^{1}

= \tan^{- 1} (e^{1}) - \tan^{- 1} (e^{0})

= \tan^{- 1} (e) - \tan^{- 1} (1)

= \tan^{- 1} (e) - \tan^{- 1} \tan \frac{π}{4}

= \tan^{- 1} (e) - \frac{π}{4}

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