Correct Answer is : log⁡(ex−1)−x+c ≤ft(e^{{x}}-1)-x+clog(ex−1)−x+c, where c is the constant of integration.

≤ft(e^{{x}}-1)+x+c, where c is the constant of integration.

≤ft(e^{{x}}-1)-x+c, where c is the constant of integration.

x- ≤ft(e^{x}-1)+c, where c is the constant of integration.

≤ft(e^{{x}}-1)-x e^{{x}}+c, where c is the constant of integration.

Correct Answer is : log⁡(ex−1)−x+c ≤ft(e^{{x}}-1)-x+clog(ex−1)−x+c, where c is the constant of integration.

Test Index

MHTCET 23 Apr 2025 Shift 1 Paper

Question : 125 of 150

Marks: +1, -0

\int \frac{d x}{e^{x}-1}=

$\log \left(e^{\mathrm{x}}-1\right)+x+c$ , where c is the constant of integration.
$\quad \log \left(e^{\mathrm{x}}-1\right)-x+c$ , where c is the constant of integration.
$\quad x-\log \left(e^{x}-1\right)+c$ , where c is the constant of integration.
$\log \left(e^{\mathrm{x}}-1\right)-x e^{\mathrm{x}}+c$ , where c is the constant of integration.

Solution:

Concept:

Use substitution followed by partial fractions to simplify the integral.

Explanation:

Let

I = \int \frac{dx}{e^x - 1}

Put

e^x - 1 = t

, so

e^x = t + 1

Differentiating,

e^x dx = dt

, therefore

dx = \frac{dt}{t+1}

Substitute into

I

I = \int \frac{dt}{t(t+1)}

Write the integrand as partial fractions:

\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}

Hence,

I = \int \left(\frac{1}{t} - \frac{1}{t+1}\right) dt

I = \log|t| - \log|t+1| + c

Now replace

t = e^x - 1

I = \log|e^x - 1| - \log|e^x| + c

Since

\log|e^x| = x

I = \log|e^x - 1| - x + c

This matches option B.

Answer:

\int \frac{dx}{e^x - 1} = \log\left|e^x - 1\right| - x + c

The correct option is B.

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