# How do you integrate f(x)=x^2/(3x-1) using the quotient rule?

Jun 13, 2018

$\frac{3 {x}^{2} - 2 x}{2 x} ^ 2$

#### Explanation:

If we have a function $f \left(x\right)$ being defined as equal to $\frac{h \left(x\right)}{g} \left(x\right)$, the Quotient Rule tells us that the derivative will be equal to

$\frac{h ' \left(x\right) g \left(x\right) - h \left(x\right) g ' \left(x\right)}{g} {\left(x\right)}^{2}$

If we define our $f \left(x\right)$ in such a way, then we know

$h \left(x\right) = {x}^{2} \implies h ' \left(x\right) = 2 x$

$g \left(x\right) = 3 x - 1 \implies g ' \left(x\right) = 3$

We have everything we need to plug in, so let's do that now!

$\frac{2 x \cdot \left(3 x - 1\right) - {x}^{2} \left(3\right)}{2 x} ^ 2$

Which simplifies to

$\frac{6 {x}^{2} - 2 x - 3 {x}^{2}}{2 {x}^{2}}$

Further simplifying to get

$f ' \left(x\right) = \overline{\underline{| \frac{3 {x}^{2} - 2 x}{2 x} ^ 2} |}$

Hope this helps!