1/(e^x-1) dx is equal to: (where C is an arbitrary constant) Option 1: -(e^x-1)^(-2)(e^x)+c Option 2: _e, e^x/(e^x-1), +c Option 3: _e, (e^x-1)/e^x, +c Option 4: e^x+c

Answer: _e, (e^x-1)/e^x, +c. Solution: Put t = e^x, so dx = dt/t and the integral becomes /(t(t-1)). Using partial fractions, 1/(t(t-1)) = 1/(t-1)-1/t. Integrating gives, (t-1)/t, +c = _e, (e^x-1)/e^x, +c.

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$\displaystyle\int \dfrac{1}{e^x-1}\,dx$ is equal to:

(where C is an arbitrary constant)

$-(e^x-1)^{-2}(e^x)+c$

$\log_e\left|\dfrac{e^x}{e^x-1}\right|+c$

$\log_e\left|\dfrac{e^x-1}{e^x}\right|+c$

$e^x+c$

✅ Correct Option: 3

Put $t = e^x$ , so $dx = \dfrac{dt}{t}$ and the integral becomes $\int\dfrac{dt}{t(t-1)}$ .

Using partial fractions, $\dfrac{1}{t(t-1)} = \dfrac{1}{t-1}-\dfrac{1}{t}$ .

Integrating gives $\log\left|\dfrac{t-1}{t}\right|+c = \log_e\left|\dfrac{e^x-1}{e^x}\right|+c$ .

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CUET Mathematics Past Year Questions (Topic-Wise)

$\displaystyle\int \dfrac{1}{e^x-1}\,dx$ is equal to:

(where C is an arbitrary constant)