# \frac{dy}{dx} = \frac{1}{x^3 + 9x}

## Question:

{eq}\frac{dy}{dx} = \frac{1}{x^3 + 9x} {/eq}

## Some Important formulas of Integration:

To solve the problem, we need to know following formulas of integration:

1.){eq}\int {\frac{{f'(x)}}{{f(x)}}dx} = \ln \left| {f(x)} \right| + c {/eq}

2.){eq}\int {\frac{{dx}}{x}} = ln(x) + c {/eq}

{eq}\eqalign{ & \frac{{dy}}{{dx}} = \frac{1}{{{x^3} + 9x}} \cr \Rightarrow& \frac{{dy}}{{dx}} = \frac{1}{{9x}} - \frac{x}{{9\left( {{x^2} + 9} \right)}} \cr \Rightarrow& dy = \frac{{dx}}{{9x}} - \frac{{2xdx}}{{18\left( {{x^2} + 9} \right)}} \cr \Rightarrow& \int {dy} = \int {\frac{{dx}}{{9x}}} - \int {\frac{{2xdx}}{{18\left( {{x^2} + 9} \right)}} + c} \cr \Rightarrow& y = \frac{1}{9}\ln \left( x \right) - \int {\frac{{d\left( {{x^2} + 9} \right)}}{{18\left( {{x^2} + 9} \right)}} + c} \cr \Rightarrow& y = \frac{1}{9}\ln \left( x \right) - \frac{1}{{18}}\ln \left[ {\left( {{x^2} + 9} \right)} \right] + c \cr} {/eq}

Where {eq}c {/eq} is the constant of integration.