Calculus Examples

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

Step 1

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

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

Step 2

Set up the integral to solve.

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

Step 3

Split the single integral into multiple integrals.

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

Step 4

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

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

Step 5

By the Power Rule, the integral of $x^{\frac{1}{2}}$ with respect to $x$ is $\frac{2}{3} x^{\frac{3}{2}}$ .

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

Step 6

Apply basic rules of exponents.

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Step 6.1

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

$\frac{2}{3} x^{\frac{3}{2}} + C + \int \frac{1}{x^{\frac{1}{2}}} d x$

Step 6.2

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

$\frac{2}{3} x^{\frac{3}{2}} + C + \int {(x^{\frac{1}{2}})}^{- 1} d x$

Step 6.3

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

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Step 6.3.1

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

$\frac{2}{3} x^{\frac{3}{2}} + C + \int x^{\frac{1}{2} \cdot - 1} d x$

Step 6.3.2

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

$\frac{2}{3} x^{\frac{3}{2}} + C + \int x^{\frac{- 1}{2}} d x$

Step 6.3.3

Move the negative in front of the fraction.

$\frac{2}{3} x^{\frac{3}{2}} + C + \int x^{- \frac{1}{2}} d x$

Step 7

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

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

Step 8

Simplify.

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

Step 9

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

$F (x) =$ $\frac{2}{3} x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + C$