Calculus Examples

$\int \frac{6}{x^{2}} - \frac{1}{\sqrt[4]{x^{3}}} + \frac{1}{\sqrt{x}} d x$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{x^{3}}$ as $x^{\frac{3}{4}}$ .

$\frac{d}{d x} [\int \frac{6}{x^{2}} - \frac{1}{x^{\frac{3}{4}}} + \frac{1}{\sqrt{x}} d x]$

Step 2

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

$\frac{d}{d x} [\int \frac{6}{x^{2}} - \frac{1}{x^{\frac{3}{4}}} + \frac{1}{x^{\frac{1}{2}}} d x]$

Step 3

$\int \frac{6}{x^{2}} - \frac{1}{x^{\frac{3}{4}}} + \frac{1}{x^{\frac{1}{2}}} d x$ is an antiderivative of $\frac{6}{x^{2}} - \frac{1}{x^{\frac{3}{4}}} + \frac{1}{x^{\frac{1}{2}}}$ , so by definition $\frac{d}{d x} [\int \frac{6}{x^{2}} - \frac{1}{x^{\frac{3}{4}}} + \frac{1}{x^{\frac{1}{2}}} d x]$ is $\frac{6}{x^{2}} - \frac{1}{x^{\frac{3}{4}}} + \frac{1}{x^{\frac{1}{2}}}$ .

$\frac{6}{x^{2}} - \frac{1}{x^{\frac{3}{4}}} + \frac{1}{x^{\frac{1}{2}}}$

Step 4

Reorder terms.

$\frac{6}{x^{2}} + \frac{1}{x^{\frac{1}{2}}} - \frac{1}{x^{\frac{3}{4}}}$