Question

$$ int \frac { 1 } { 2 sqrt { x } } d x $$

Explanation:

Step1: Rewrite the integrand

Rewrite \(\frac{1}{2\sqrt{x}}\) as \(\frac{1}{2}x^{-\frac{1}{2}}\). So the integral becomes \(\int\frac{1}{2}x^{-\frac{1}{2}}dx\).

Step2: Apply the power rule for integration

The power rule for integration is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)). Here \(n=-\frac{1}{2}\), and \(\int\frac{1}{2}x^{-\frac{1}{2}}dx=\frac{1}{2}\int x^{-\frac{1}{2}}dx\).
Using the power rule: \(\frac{1}{2}\times\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C\).

Step3: Simplify the expression

Simplify the exponent and the denominator: \(-\frac{1}{2}+1=\frac{1}{2}\). Then \(\frac{1}{2}\times\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C=x^{\frac{1}{2}}+C=\sqrt{x}+C\).

Answer:

\(\sqrt{x}+C\)