Question

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

Explanation:

Step1: Expand the numerator

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

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 $\int\frac{1}{2}\sqrt{x}dx=\frac{1}{2}\int x^{\frac{1}{2}}dx=\frac{1}{2}\times\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\frac{1}{2}\times\frac{x^{\frac{3}{2}}}{\frac{3}{2}}=\frac{1}{3}x^{\frac{3}{2}}$.
For $\int2dx=2x$.
For $\int2x^{-\frac{1}{2}}dx=2\times\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2\times\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 4\sqrt{x}$.

Step3: Combine the results

The result of the integral is $\frac{1}{3}x^{\frac{3}{2}}+2x + 4\sqrt{x}+C$.

Answer:

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