Find the most general antiderivative G(x) of the function g(x) = \left(x + \frac{1}{\sqrt x}...

Question:

Find the most general antiderivative {eq}G(x) {/eq} of the function {eq}g(x) = \left( x + \frac{1}{\sqrt x} \right)^2 {/eq}

Indefinite Integral in Calculus:

Indefinite integral is a process of finding out the antiderivative of a function. The indefinite integral for a given function {eq}g {/eq} of a real variable {eq}x {/eq} is {eq}\displaystyle \int g\left( x\right) dx . {/eq}

To compute an indefinite integral, we can take help of common integrals to make the process easier. In this problem,we'll use the integral power rule {eq}\displaystyle \int x^n \ dx =\dfrac{x^{n+1}}{n+1}+C. {/eq} and the common integral {eq}\int \dfrac{1}{x} \ dx =\ln x + C. {/eq}

Answer and Explanation:

We need to find out the antiderivative of {eq}\displaystyle g(x) = \left(x + \frac{1}{\sqrt x} \right)^2 {/eq}.

Hence, we need to compute {eq}\displaystyle \int \left(x + \frac{1}{\sqrt x} \right)^2 \ dx {/eq}


Use the algebraic square rule:

{eq}=\displaystyle \int \left(x^2 +2 x \frac{1}{\sqrt x} + \dfrac{1}{x} \right) \ dx {/eq}

{eq}=\displaystyle \int \left(x^2 +2 x^{\frac{1}{2}} + \dfrac{1}{x} \right) \ dx {/eq}


Apply integral sum rule:

{eq}= \displaystyle \int x^2 \ dx +2\int x^{\frac{1}{2}} \ dx +\int \dfrac{1}{x} \ dx {/eq}

{eq}=\displaystyle \dfrac{x^{2+1}}{2+1}+ \dfrac{2x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\ln x +C \quad \text{where C is an arbitrary constant} {/eq}

{eq}=\displaystyle \dfrac{x^{3}}{3}+ \dfrac{4x^{\frac{3}{2}}}{3}+\ln x +C {/eq}


Therefore the antiderivative is {eq}{\boxed{\displaystyle \int \left(x + \frac{1}{\sqrt x} \right)^2 \ dx = \dfrac{x^{3}}{3}+ \dfrac{4x^{\frac{3}{2}}}{3}+\ln x +C . }} {/eq}


Learn more about this topic:

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Antiderivative: Rules, Formula & Examples

from Calculus: Help and Review

Chapter 8 / Lesson 12
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