Answer :
Sure, let's solve the integral [tex]\(\int \left( \sqrt{x} + \frac{1}{2 \sqrt{x}} \right) dx \)[/tex] step by step.
First, we will break the integral into two separate integrals:
[tex]\[ \int \left( \sqrt{x} + \frac{1}{2 \sqrt{x}} \right) dx = \int \sqrt{x} \, dx + \int \frac{1}{2 \sqrt{x}} \, dx \][/tex]
Now we'll solve each integral separately.
### Integral 1: [tex]\(\int \sqrt{x} \, dx\)[/tex]
Recall that [tex]\(\sqrt{x}\)[/tex] is the same as [tex]\(x^{1/2}\)[/tex]. Therefore, we can rewrite the first integral as:
[tex]\[ \int x^{1/2} \, dx \][/tex]
We use the power rule for integration which states that [tex]\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)[/tex]. Here, [tex]\(n = \frac{1}{2}\)[/tex]:
[tex]\[ \int x^{1/2} \, dx = \frac{x^{(1/2)+1}}{(1/2)+1} = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \][/tex]
So, the result for this part is:
[tex]\[ \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} \][/tex]
### Integral 2: [tex]\(\int \frac{1}{2 \sqrt{x}} \, dx\)[/tex]
Recall that [tex]\(\sqrt{x}\)[/tex] is the same as [tex]\(x^{1/2}\)[/tex], so [tex]\(\frac{1}{2 \sqrt{x}}\)[/tex] can be written as [tex]\(\frac{1}{2} x^{-1/2}\)[/tex]. Therefore, we can rewrite the second integral as:
[tex]\[ \int \frac{1}{2} x^{-1/2} \, dx \][/tex]
We can factor out the constant [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \frac{1}{2} \int x^{-1/2} \, dx \][/tex]
Using the power rule for integration where [tex]\(n = -\frac{1}{2}\)[/tex]:
[tex]\[ \int x^{-1/2} \, dx = \frac{x^{(-1/2)+1}}{(-1/2)+1} = \frac{x^{1/2}}{1/2} = 2 x^{1/2} \][/tex]
Therefore:
[tex]\[ \frac{1}{2} \int x^{-1/2} \, dx = \frac{1}{2} \cdot 2 x^{1/2} = x^{1/2} = \sqrt{x} \][/tex]
### Combine the results
Now we combine the two results:
[tex]\[ \int \left( \sqrt{x} + \frac{1}{2 \sqrt{x}} \right) dx = \int \sqrt{x} \, dx + \int \frac{1}{2 \sqrt{x}} \, dx \][/tex]
[tex]\[ = \frac{2}{3} x^{3/2} + \sqrt{x} + C \][/tex]
Therefore, the solution to the integral [tex]\(\int \left( \sqrt{x} + \frac{1}{2 \sqrt{x}} \right) dx \)[/tex] is:
[tex]\[ \boxed{\frac{2}{3} x^{3/2} + \sqrt{x} + C} \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
First, we will break the integral into two separate integrals:
[tex]\[ \int \left( \sqrt{x} + \frac{1}{2 \sqrt{x}} \right) dx = \int \sqrt{x} \, dx + \int \frac{1}{2 \sqrt{x}} \, dx \][/tex]
Now we'll solve each integral separately.
### Integral 1: [tex]\(\int \sqrt{x} \, dx\)[/tex]
Recall that [tex]\(\sqrt{x}\)[/tex] is the same as [tex]\(x^{1/2}\)[/tex]. Therefore, we can rewrite the first integral as:
[tex]\[ \int x^{1/2} \, dx \][/tex]
We use the power rule for integration which states that [tex]\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)[/tex]. Here, [tex]\(n = \frac{1}{2}\)[/tex]:
[tex]\[ \int x^{1/2} \, dx = \frac{x^{(1/2)+1}}{(1/2)+1} = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \][/tex]
So, the result for this part is:
[tex]\[ \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} \][/tex]
### Integral 2: [tex]\(\int \frac{1}{2 \sqrt{x}} \, dx\)[/tex]
Recall that [tex]\(\sqrt{x}\)[/tex] is the same as [tex]\(x^{1/2}\)[/tex], so [tex]\(\frac{1}{2 \sqrt{x}}\)[/tex] can be written as [tex]\(\frac{1}{2} x^{-1/2}\)[/tex]. Therefore, we can rewrite the second integral as:
[tex]\[ \int \frac{1}{2} x^{-1/2} \, dx \][/tex]
We can factor out the constant [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \frac{1}{2} \int x^{-1/2} \, dx \][/tex]
Using the power rule for integration where [tex]\(n = -\frac{1}{2}\)[/tex]:
[tex]\[ \int x^{-1/2} \, dx = \frac{x^{(-1/2)+1}}{(-1/2)+1} = \frac{x^{1/2}}{1/2} = 2 x^{1/2} \][/tex]
Therefore:
[tex]\[ \frac{1}{2} \int x^{-1/2} \, dx = \frac{1}{2} \cdot 2 x^{1/2} = x^{1/2} = \sqrt{x} \][/tex]
### Combine the results
Now we combine the two results:
[tex]\[ \int \left( \sqrt{x} + \frac{1}{2 \sqrt{x}} \right) dx = \int \sqrt{x} \, dx + \int \frac{1}{2 \sqrt{x}} \, dx \][/tex]
[tex]\[ = \frac{2}{3} x^{3/2} + \sqrt{x} + C \][/tex]
Therefore, the solution to the integral [tex]\(\int \left( \sqrt{x} + \frac{1}{2 \sqrt{x}} \right) dx \)[/tex] is:
[tex]\[ \boxed{\frac{2}{3} x^{3/2} + \sqrt{x} + C} \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.