∫ 0 4 1 1 + x d x = ________

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Question: Answered & Verified by Expert

\int_{0}^{4} \frac{1}{1 + \sqrt{x}} d x =

________

MathematicsDefinite IntegrationMHT CETMHT CET 2019 (02 May Shift 1)

Options:

A $\log (\frac{e^{4}}{6})$
B $\log (\frac{e^{4}}{3})$
C $\log (\frac{e^{4}}{9})$
D $\log (\frac{e^{3}}{4})$

Solution:

2100 Upvotes Verified Answer

The correct answer is:

\log (\frac{e^{4}}{9})

Let

I = \int_{0}^{4} \frac{d x}{1 + \sqrt{x}}

put

x = t^{2}

d x = 2 t d t

Then,

I = \int_{0}^{2} \frac{2 t d t}{1 + t} = 2 \int_{0}^{2} \frac{(1 + t) - 1}{1 + t} d t

I = 2 \int_{0}^{2} (1 - \frac{1}{1 + t}) d t

I = 2 (2 - l n 3) = l o g (\frac{e^{4}}{9})

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