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UPSC NDA 2 2022 Mathematics Solved Paper

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Question : 94 of 120

Marks: +1, -0

What is

\displaystyle\int_0^1 \ln\left(\frac{1}{x}-1\right)

dx equal to ?

$-1$
0
1
ln 2

Solution:

Concept:

The integral simplifies using

\ln(a/b) = \ln a - \ln b

and the property

\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx

.

Explanation:

Start with

I = \int_0^1 \ln\left(\frac{1}{x}-1\right) dx = \int_0^1 \ln\left(\frac{1-x}{x}\right) dx

.

Using

\ln(a/b) = \ln a - \ln b

, we get

I = \int_0^1 \ln(1-x)\,dx - \int_0^1 \ln x\,dx

.

Apply the property

\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx

to the second integral:

\int_0^1 \ln x\,dx = \int_0^1 \ln(1-x)\,dx

.

Thus

I = \int_0^1 \ln(1-x)\,dx - \int_0^1 \ln(1-x)\,dx = 0

.

Answer:

0 (Option B).

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