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UPSC NDA II 2015 Mathematics Paper

Question : 69 of 120

Marks: +1, -0

\smallint \frac{dx}{1+e^{-x}}

Where c is the constant of integration

Solution:

Concept:

The integral

\int\frac{dx}{1+e^{-x}}

is solved by rewriting the integrand and using the substitution

t = 1+e^{x}

with the formula

\int\frac{dt}{t} = \ln|t| + c

Explanation:

Start with

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

Rewrite

e^{-x}

\frac{1}{e^{x}}

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

Let

t = 1+e^{x}

Then

dt = e^{x}dx

Substitute:

I = \int\frac{dt}{t} = \ln|t| + c

Replace

t

with

1+e^{x}

I = \ln(1+e^{x}) + c

Thus the integral equals

\ln(1+e^{x}) + c

Answer:

\ln(1+e^{x}) + c

(option C).

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