Question

evaluate the following indefinite integral.

$$ int \frac { 7 } { sqrt { x } } d x $$
$$ int \frac { 7 } { sqrt { x } } d x = square $$

Explanation:

Step1: Rewrite the integrand

We know that \(\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}\). So, \(\int\frac{7}{\sqrt{x}}dx = 7\int x^{-\frac{1}{2}}dx\)

Step2: Apply the power - rule for integration

The power - rule for integration is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)). For \(n=-\frac{1}{2}\), we have:
\(7\int x^{-\frac{1}{2}}dx=7\times\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C\)
\(=7\times\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C\)

Step3: Simplify the expression

\(7\times\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C=14\sqrt{x}+C\)

Answer:

\(14\sqrt{x}+C\)