# Find the antiderivative of ..?

## $\frac{6 {x}^{5} - 7 {x}^{3} + 12 x}{2 x}$

Jan 25, 2018

$\int \frac{6 {x}^{5} - 7 {x}^{3} + 12 x}{2 x} \mathrm{dx} = \frac{6 {x}^{5}}{10} - \frac{7 {x}^{3}}{6} + 6 x + \text{C}$

#### Explanation:

Given: $\int \frac{6 {x}^{5} - 7 {x}^{3} + 12 x}{2 x} \mathrm{dx}$

Take out the constant $\frac{1}{2}$

$\frac{1}{2} \cdot \int \frac{6 {x}^{5} - 7 {x}^{3} + 12 x}{x} \mathrm{dx}$

We can separate each term and integrate each term individually so we can rewrite the integral as:

$\frac{1}{2} \left[\int \frac{6 {x}^{5}}{x} \mathrm{dx} - \int \frac{7 {x}^{3}}{x} \mathrm{dx} + \int \frac{12 x}{x} \mathrm{dx}\right]$

But before we integrate, simplify!

$\frac{1}{2} \left[\int 6 {x}^{4} \mathrm{dx} - \int 7 {x}^{2} \mathrm{dx} + \int 12 \mathrm{dx}\right]$

Now we can integrate each term:

$\int 6 {x}^{4} \mathrm{dx} \implies \int c {x}^{a} \mathrm{dx} = \frac{c {x}^{a + 1}}{a + 1} \implies \frac{6 {x}^{4 + 1}}{4 + 1} = \frac{6 {x}^{5}}{5}$

$- \int 7 {x}^{2} \mathrm{dx} \implies \int c {x}^{a} \mathrm{dx} = \frac{c {x}^{a + 1}}{a + 1} \implies \frac{7 {x}^{2 + 1}}{2 + 1} = - \frac{7 {x}^{3}}{3}$

$\int 12 \mathrm{dx} \implies \int c \mathrm{dx} = c x \implies = 12 x$

Putting it all together we get

$= \frac{1}{2} \left[\frac{6 {x}^{5}}{5} - \frac{7 {x}^{3}}{3} + 6 x\right]$

$= \frac{6 {x}^{5}}{10} - \frac{7 {x}^{3}}{6} + 6 x + \text{C} \leftarrow$ Don't Forget the Constant!