# Find the function whose derivative is given ?

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

Mar 21, 2017

$y = \frac{1}{8} {\left(2 {x}^{3} + {x}^{2}\right)}^{4} + C$

We must be given a point on the curve to determine the value of C.

#### Explanation:

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

Multiply both sides by $\mathrm{dx}$:

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

Integrate both sides:

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

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

Let $u = 2 {x}^{3} + {x}^{2}$, then $\mathrm{du} = \left(6 {x}^{2} + 2 x\right) \mathrm{dx}$

This makes the integral become:

$y = \frac{1}{2} \int {u}^{3} \mathrm{du}$

$y = \left(\frac{1}{2}\right) \left(\frac{1}{4}\right) {u}^{4} + C$

$y = \frac{1}{8} {\left(2 {x}^{3} + {x}^{2}\right)}^{4} + C$