Mathematics

$$\int_0^1 {\dfrac{{dx}}{{x + \sqrt x }}} $$ equals


ANSWER

$$2ln\ 2$$


SOLUTION

Consider the given integral.

$$I=\int_{0}^{1}{\dfrac{dx}{x+\sqrt{x}}}$$

$$I=\int_{0}^{1}{\dfrac{dx}{\sqrt{x}\left( 1+\sqrt{x} \right)}}$$

 

Let $$t=1+\sqrt{x}$$

$$ \dfrac{dt}{dx}=\dfrac{1}{2\sqrt{x}} $$

$$ 2dt=\dfrac{1}{\sqrt{x}}dx $$

 

Therefore,

$$ I=2\int_{1}^{2}{\dfrac{dt}{t}} $$

$$ I=2\left[ \ln \left( t \right) \right]_{1}^{2} $$

$$ I=2\left[ \ln \left( 2 \right)-\ln \left( 1 \right) \right] $$

$$ I=2\ln \left( 2 \right) $$

 

Hence, the value of integral is $$2\ln \left( 2 \right)$$.

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