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MHTCET 23 Apr 2025 Shift 2 Paper

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Question : 129 of 150

Marks: +1, -0

\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x=

$\frac{1}{2}$
1
2
0

Solution:

Concept:

Use the definite integral property

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

and simplify the logarithm.

Explanation:

Let

I = \int_{0}^{1} \log\left(\frac{1}{x}-1\right) dx

.

Rewrite the integrand:

\frac{1}{x}-1 = \frac{1-x}{x}

.

So,

I = \int_{0}^{1} \log\left(\frac{1-x}{x}\right) dx

.

Now apply the property with

a = 1

: replace

x

by

1-x

.

Then

I = \int_{0}^{1} \log\left(\frac{x}{1-x}\right) dx

.

Add the two expressions for

I

:

2I = \int_{0}^{1} \left[\log\left(\frac{1-x}{x}\right) + \log\left(\frac{x}{1-x}\right)\right] dx

Using

\log A + \log B = \log(AB)

, the sum inside becomes

\log(1) = 0

.

Therefore,

2I = 0

, so

I = 0

.

Answer:

I = 0

, which matches option D.

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