If ∫dx1+ex=x+Klog⁡1+ex+C then K is equal to

Substituting 1+ex=t we have∫dx1+ex=∫dtt(t−1)=∫1t−1dt−∫dtt=log⁡[t−1]−log⁡t+C=log⁡ex−log⁡1+ex+C=x−log⁡1+ex+C

If ∫dx1+ex=x+Klog⁡1+ex+C then K is equal to

Substituting 1+ex=t we have∫dx1+ex=∫dtt(t−1)=∫1t−1dt−∫dtt=log⁡[t−1]−log⁡t+C=log⁡ex−log⁡1+ex+C=x−log⁡1+ex+C

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If $\int \frac{d x}{1 + e^{x}} = x + Klog (1 + e^{x}) + C$ then $K$ is equal to

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Substituting $1 + e^{x} = t$ we have

$\begin{array}{l} \int \frac{d x}{1 + e^{x}} = \int \frac{d t}{t (t - 1)} = \int \frac{1}{t - 1} d t - \int \frac{d t}{t} \\ = \log [t - 1] - \log t + C \\ = \log e^{x} - \log (1 + e^{x}) + C \\ = x - \log (1 + e^{x}) + C \end{array}$

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If ∫dx1+ex=x+Klog⁡1+ex+C then K is equal to

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If $\int \frac{d x}{1 + e^{x}} = x + Klog (1 + e^{x}) + C$ then $K$ is equal to