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

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

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

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

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MHTCET 25 Apr 2025 Shift 1 Paper

Question : 111 of 150

Marks: +1, -0

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

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

Solution:

Concept:

Rewrite the integrand by adding and subtracting

e^x

to simplify it into standard known integrals.

Explanation:

Start with the given integral:

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

Add and subtract

e^x

in the numerator:

\frac{1}{e^{x}+1} = \frac{1+e^x - e^x}{e^{x}+1} = 1 - \frac{e^x}{e^{x}+1}

So the integral becomes:

I = \int \left(1 - \frac{e^x}{e^{x}+1}\right) dx

Integrate term by term:

\int 1 \, dx = x

For the second term, let

t = e^x + 1

. Then

dt = e^x dx

, so:

\int \frac{e^x}{e^{x}+1} dx = \int \frac{dt}{t} = \log|t| = \log(e^x+1)

Therefore:

I = x - \log(e^x+1) + c

Since

e^x+1

is always positive, the absolute value is not needed.

Answer:

x - \log\left(e^x+1\right) + c

This matches option B.

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