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Question

# Find the most general antiderivative of the function. (Check your answer by differentiation.)$f(x) = 2√x+6cosx$

Solution

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We will find the most general antiderivative of the function $f(x)=2\sqrt{x} + 6\cos{x}$. It will be denoted by $F$. Therefore, we will find the function for which $F^{\prime}=f$. The first summand is

$2\sqrt{x} = 2x^{\frac{1}{2}}$

so its antiderivative by using the inverse of Power Rule for derivative is

$2x^{\frac{1}{2}} = \left(2\cdot \frac{2}{3} \cdot x^{\frac{3}{2}}\right)^{\prime} = \left(\frac{4}{3} \cdot x^{\frac{3}{2}}\right)^{\prime}$

The antiderivative of other summand is

$6\cos{x} = \left(6\sin(x)\right)^{\prime}$

Therefore, the solution is

$\boxed{\frac{4}{3} \cdot x^{\frac{3}{2}} + 6\sin(x) + C}$

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