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Question

Find the most general antiderivative of the function. (Check your answer by differentiation.)

f(x)=2x+6cosxf(x) = 2√x+6cosx

Solution

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

2x=2x122\sqrt{x} = 2x^{\frac{1}{2}}

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

2x12=(223x32)=(43x32)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

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

Therefore, the solution is

43x32+6sin(x)+C\boxed{\frac{4}{3} \cdot x^{\frac{3}{2}} + 6\sin(x) + C}

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