The correct option is B 2 log 2 ∫_0^1dx(x+√(x)) #10; x=t^2 #10; ∫_0^12t #160; dtt^2+ #10;t=2∫_0^1dtt+1=2 log (t+1)∫_0^1 #10; =2 ...

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Integration of Trigonometric Functions

∫01dx/x+√x=

Question

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

log 2

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2 log 2

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3 log 3

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\frac{1}{2}

log2

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Solution

The correct option is B 2 log 2
$\int_{0}^{1} \frac{d x}{(x + \sqrt{x})}$
$x = t^{2}$
$\int_{0}^{1} \frac{2 t d t}{t^{2} + t} = 2 \int_{0}^{1} \frac{d t}{t + 1} = 2 l o g (t + 1) \int_{0}^{1}$
$= 2 l o g (2) - 2 l o g (1)$
$= 2 l o g 2$

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