MHTCET 5 May 2025 Shift 2 Paper

Question : 126 of 150

Marks: +1, -0

\int_{0}^{1} \log (x+1) d x=

Solution:

Concept:

Use integration by parts to evaluate the definite integral of a logarithmic function.

Explanation:

Let

I = \int_{0}^{1} \log (x+1) \, dx

Applying integration by parts, treat

\log(x+1)

as the first function and

1

as the second function.

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

Simplify the remaining integral:

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

So,

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

Evaluate the boundary term:

\left[ x \log(x+1) \right]_{0}^{1} = 1 \cdot \log 2 - 0 = \log 2

Evaluate the second integral:

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

Thus,

I = \log 2 - (1 - \log 2) = 2\log 2 - 1

Answer:

2\log 2 - 1

, which is option D.

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