What is ∫{dx}{2^{x}-1} equal to?

Correct Answer is : ln(1−2−x)ln2+c{ln(1-2^{-x})}{ln2}+cln2ln(1−2−x)+c

Correct Answer is : ln(1−2−x)ln2+c{ln(1-2^{-x})}{ln2}+cln2ln(1−2−x)+c

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UPSC NDA I 2018 Mathematics Solved Paper

Question : 26 of 120

Marks: +1, -0

What is

∫\frac{dx}{2^{x}-1}

equal to?

Solution:

Concept:

Integration by substitution using the derivative of exponential functions.

Explanation:

We need to evaluate

∫\frac{dx}{2^{x}-1}

Rewrite the integrand by multiplying numerator and denominator by

2^{-x}

∫\frac{dx}{2^{x}-1} = ∫\frac{2^{-x}dx}{1-2^{-x}}

Let

t = 1 - 2^{-x}

Differentiate both sides:

\frac{dt}{dx} = 2^{-x}\ln 2

, so

2^{-x}dx = \frac{dt}{\ln 2}

Substitute into the integral:

∫\frac{2^{-x}dx}{1-2^{-x}} = ∫\frac{1}{t} \cdot \frac{dt}{\ln 2} = \frac{1}{\ln 2} ∫\frac{dt}{t}

Integrate:

\frac{1}{\ln 2} \ln |t| + c = \frac{\ln(1-2^{-x})}{\ln 2} + c

Answer:

Option B:

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

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