₀¹ {1}{2+{x}} d x=

Correct Answer is : 2log⁡(4e9)2 ≤ft(4 e/9)2log(94e)

Correct Answer is : 2log⁡(4e9)2 ≤ft(4 e/9)2log(94e)

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MHTCET 25 Apr 2025 Shift 1 Paper

Question : 147 of 150

Marks: +1, -0

\int_{0}^{1} \frac{1}{2+\sqrt{x}} d x=

Solution:

Concept:

Use the substitution

t = 2 + \sqrt{x}

to transform the integral into a simple logarithmic form.

Explanation:

Let

I = \int_{0}^{1} \frac{1}{2+\sqrt{x}} dx

Put

t = 2 + \sqrt{x}

, so

\frac{dt}{dx} = \frac{1}{2\sqrt{x}}

Thus

dx = 2\sqrt{x}\, dt = 2(t-2)\, dt

When

x = 0

t = 2

; and when

x = 1

t = 3

The integral becomes

I = \int_{2}^{3} \frac{2(t-2)}{t} dt = 2\int_{2}^{3} \left(1 - \frac{2}{t}\right) dt

Integrating term by term gives

I = 2\left[t - 2\log t\right]_{2}^{3}

Substituting the limits,

I = 2\left[(3 - 2\log 3) - (2 - 2\log 2)\right]

This simplifies to

I = 2\left[1 - 2\log\frac{3}{2}\right] = 2 - 4\log\frac{3}{2}

Now

2 = 2\log e

and

-4\log\frac{3}{2} = 4\log\frac{2}{3} = 2\log\frac{4}{9}

Therefore

I = 2\log e + 2\log\frac{4}{9} = 2\log\left(\frac{4e}{9}\right)

Answer:

2\log\left(\frac{4e}{9}\right)

, which matches option B.

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