# What is the antiderivative of  (x+1)(2x-1)?

Mar 24, 2018

$\int \left(x + 1\right) \left(2 x - 1\right) \mathrm{dx} = \frac{2}{3} {x}^{3} + \frac{1}{2} {x}^{2} - x + \text{c}$

#### Explanation:

Find an antiderivative of a function is the same as finding the integral.

So we want to find $\int \left(x + 1\right) \left(2 x - 1\right) \mathrm{dx} = \int 2 {x}^{2} + x - 1 \mathrm{dx}$

To find this, we use the power rule

$\int {x}^{n} \mathrm{dx} = \frac{1}{n + 1} {x}^{n + 1} + \text{c}$

$\Rightarrow \int 2 {x}^{2} + x - 1 \mathrm{dx} = \frac{2}{3} {x}^{3} + \frac{1}{2} {x}^{2} - x + \text{c}$

Mar 24, 2018

$\implies \left(\frac{2}{3} {x}^{2} + \frac{1}{2} x - 1\right) x + C$

where $C$ is an arbitrary constant from the integration.

#### Explanation:

$\setminus \int \left(x + 1\right) \left(2 x - 1\right) \mathrm{dx}$

We multiply the two binomials:

$= \setminus \int \left(2 {x}^{2} + 2 x - x - 1\right) \mathrm{dx}$

$= \setminus \int \left(2 {x}^{2} + x - 1\right) \mathrm{dx}$

We break up the integral into three separate integrals:

$= \setminus \int 2 {x}^{2} \mathrm{dx} + \setminus \int x \mathrm{dx} + \setminus \int - \mathrm{dx}$

$= \textcolor{b l u e}{2 \int {x}^{2} \mathrm{dx}} + \textcolor{\mathmr{and} a n \ge}{\int x \mathrm{dx}} - \textcolor{g r e e n}{\int \mathrm{dx}}$

Now we integrate:

$= \textcolor{b l u e}{2 \left(\frac{1}{3} {x}^{3}\right)} + \textcolor{\mathmr{and} a n \ge}{\frac{1}{2} {x}^{2}} - \textcolor{g r e e n}{x} + C$

Simplifying:

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

$= \left(\frac{2}{3} {x}^{2} + \frac{1}{2} x - 1\right) x + C$

where $C$ is an arbitrary constant from the integration.