Find the Anti-Derivative 1/( square root of x)

$\frac{1}{\sqrt{x}}$

Write $\frac{1}{\sqrt{x}}$ as a function.

$f (x) = \frac{1}{\sqrt{x}}$

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Set up the integral to solve.

$F (x) = \int \frac{1}{\sqrt{x}} d x$

Apply basic rules of exponents.

Tap for more steps…

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

Move $x^{\frac{1}{2}}$ out of the denominator by raising it to the $- 1$ power.

$\int {(x^{\frac{1}{2}})}^{- 1} d x$

Multiply the exponents in ${(x^{\frac{1}{2}})}^{- 1}$ .

Tap for more steps…

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int x^{\frac{1}{2} \cdot - 1} d x$

Combine $\frac{1}{2}$ and $- 1$ .

$\int x^{\frac{- 1}{2}} d x$

Move the negative in front of the fraction.

$\int x^{- \frac{1}{2}} d x$

By the Power Rule, the integral of $x^{- \frac{1}{2}}$ with respect to $x$ is $2 x^{\frac{1}{2}}$ .

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

The answer is the antiderivative of the function $f (x) = \frac{1}{\sqrt{x}}$ .

$F (x) =$ $2 x^{\frac{1}{2}} + C$

Find the Anti-Derivative 1/( square root of x)

Download our
App from the store