Test Index

UPSC NDA I 2022 Mathematics Solved Paper

Question : 63 of 120

Marks: +1, -0

I_1 = \int \frac{e^x dx}{e^x + e^{-x}}

I_2 = \int \frac{dx}{e^{2x} + 1}

I_{1} + I_{2}

equal to?

Solution:

Concept:

Adding two integrals after algebraic manipulation simplifies to a basic integral.

Explanation:

Rewrite

I_1

: multiply numerator and denominator of

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

e^x

to obtain

\frac{e^{2x}}{e^{2x}+1}

Thus

I_1 = \int \frac{e^{2x}}{e^{2x}+1} dx

I_2 = \int \frac{1}{e^{2x}+1} dx

Add the two integrals:

I_1 + I_2 = \int \left( \frac{e^{2x}}{e^{2x}+1} + \frac{1}{e^{2x}+1} \right) dx

Combine the numerators:

\frac{e^{2x}+1}{e^{2x}+1} = 1

I_1 + I_2 = \int 1 \, dx = x + C

Answer:

I_1 + I_2 = x + C

(Option B).

Go to Question:

Prev Question

Next Question