# How do you find \int ( 2x + \frac { 3} { x ^ { 2} } ) d x?

Feb 5, 2017

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

#### Explanation:

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

We can say

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

Also since $\int c x \mathrm{dx} = c \int x \mathrm{dx}$

we can say

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

Then

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