# How do you evaluate the indefinite integral int (x^2-x+5)dx?

Jan 22, 2017

$\int \left({x}^{2} - x + 5\right) \mathrm{dx} = {x}^{3} / 3 - {x}^{2} / 2 + 5 x + C$

#### Explanation:

Remember that an indefinite integral is an antiderivative, and since the derivative of sums is the sum of derivatives then

$\int f \left(x\right) + g \left(x\right) + h \left(x\right) \mathrm{dx} = \int f \left(x\right) \mathrm{dx} + \int g \left(x\right) \mathrm{dx} + \int h \left(x\right) \mathrm{dx}$

it is the same for integrals

In this case

$f \left(x\right) = {x}^{2}$

$g \left(x\right) = - x$

$h \left(x\right) = 5$

Since $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$

$\int f \left(x\right) \mathrm{dx} = \int {x}^{2} \mathrm{dx} = {x}^{3} / 3$

and also since $\int - x \mathrm{dx} = - \int x \mathrm{dx}$

$\int g \left(x\right) \mathrm{dx} = \int \left(- x\right) \mathrm{dx} = - \int x \mathrm{dx} = - {x}^{2} / 2$

and since for a constant $c$, $\int c \mathrm{dx} = c x$ then

$\int h \left(x\right) \mathrm{dx} = \int 5 \mathrm{dx} = 5 x$

Then put them together

$\int \left({x}^{2} - x + 5\right) \mathrm{dx} = {x}^{3} / 3 - {x}^{2} / 2 + 5 x + C$

Jan 22, 2017

I tried this:

#### Explanation:

We can break it into three parts and write:
$\int {x}^{2} \mathrm{dx} - \int x \mathrm{dx} + \int 5 \mathrm{dx} =$

we now use the general integration formula as:
$\textcolor{red}{\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + c}$

we can write our integral as:
$\int {x}^{2} \mathrm{dx} - \int x \mathrm{dx} + 5 \int {x}^{0} \mathrm{dx} =$
and we get:
$= {x}^{3} / 3 - {x}^{2} / 2 + 5 x + c$