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  • Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 6

Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 6

Answers (1)

best_answer

Answer:

\log 2
Given:

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

Hint:

Explanation:  

Let

\begin{aligned} &I=\int_{0}^{\infty} \frac{1}{1+e^{x}} d x \\\\ &=\int_{0}^{\infty} \frac{1}{e^{x}\left(\frac{1}{e^{x}}+1\right)} d x \\\\ &=\int_{0}^{\infty} \frac{e^{-x}}{1+e^{-x}} d x \end{aligned}

Put

\begin{aligned} &1+e^{-x}=t \\\\ &-e^{-x} d x=d t \\\\ &e^{-x} d x=-d t \\\\ &=\int_{0}^{\infty} \frac{-d t}{t} \\\\ &=[-\log |t|]_{0}^{\infty} \end{aligned}

\begin{aligned} &=\left[-\log \left|1+e^{-x}\right|\right]_{0}^{\infty} \\\\ &=\left[-\log \left|1+e^{-\infty}\right|\right]-\left[-\log \left|1+e^{-0}\right|\right] \end{aligned}

\begin{aligned} &=[-\log |1+0|]-[-\log |1+1|],\left[\therefore e^{-\infty}=0, \quad e^{-0}=1\right] \\\\ &=-[0-\log 2] \\\\ &=\log 2 \end{aligned}

 

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