dx-e-2x-1-

Question Number 144716 by mathdanisur last updated on 28/Jun/21

\int (\frac{d x}{e^{2 x} + 1}) = ?

Answered by imjagoll last updated on 28/Jun/21

let e^{x} = μ \Rightarrow dx = \frac{d μ}{e^{x}} = \frac{d μ}{μ}

I = \int (\frac{1}{μ^{2} + 1}) (\frac{d μ}{μ})

\frac{1}{μ (μ^{2} + 1)} = \frac{a}{μ} + \frac{b μ + c}{μ^{2} + 1}

1 = (a + b) μ^{2} + c μ + a

\to μ = 0 \to a = 1

μ = 1 \to 1 = 1 + b + c + 1

\to b + c = - 1

μ = - 1 \to 1 = 1 + b - c + 1

\to b - c = - 1 {\begin{cases} b = - 1 \\ c = 0 \end{cases}

I = \int (\frac{1}{μ} - \frac{μ}{μ^{2} + 1}) d μ

I = \ln μ - \frac{1}{2} \int \frac{d (μ^{2} + 1)}{μ^{2} + 1}

I = \ln μ - \frac{1}{2} \ln (μ^{2} + 1) + c

I = \ln e^{x} - \frac{1}{2} \ln (e^{2 x} + 1) + c

I = x - \frac{1}{2} \ln (e^{2 x} + 1) + c

Commented by mathdanisur last updated on 28/Jun/21

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