Question

\\(\int\frac{(\sqrt{x}+2)^{2}}{2\sqrt{x}}dx\\)

Explanation:

Step1: Expand the numerator

First, expand $(\sqrt{x}+2)^2$ using the formula $(a + b)^2=a^{2}+2ab + b^{2}$. Here $a=\sqrt{x}$ and $b = 2$, so $(\sqrt{x}+2)^2=x + 4\sqrt{x}+4$. Then the integral becomes $\int\frac{x + 4\sqrt{x}+4}{2\sqrt{x}}dx=\frac{1}{2}\int(\frac{x}{\sqrt{x}}+\frac{4\sqrt{x}}{\sqrt{x}}+\frac{4}{\sqrt{x}})dx=\frac{1}{2}\int(x^{\frac{1}{2}}+4 + 4x^{-\frac{1}{2}})dx$.

Step2: Integrate term - by - term

Use the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$).
For $\int x^{\frac{1}{2}}dx=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\frac{2}{3}x^{\frac{3}{2}}$, $\int4dx=4x$, and $\int4x^{-\frac{1}{2}}dx=4\times\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=8x^{\frac{1}{2}}$.
So $\frac{1}{2}\int(x^{\frac{1}{2}}+4 + 4x^{-\frac{1}{2}})dx=\frac{1}{2}(\frac{2}{3}x^{\frac{3}{2}}+4x + 8x^{\frac{1}{2}})+C$.

Step3: Simplify the result

$\frac{1}{2}(\frac{2}{3}x^{\frac{3}{2}}+4x + 8x^{\frac{1}{2}})+C=\frac{1}{3}x^{\frac{3}{2}}+2x + 4x^{\frac{1}{2}}+C$.

Answer:

$\frac{1}{3}x^{\frac{3}{2}}+2x + 4x^{\frac{1}{2}}+C$