Question

Find the function whose derivative is given ?

`dy/dx = (3x^2 + x)(2x^3 + x^2)^3`

Answer 1

`y = 1/8(2x^3+x^2)^4+ C`

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

Explanation:

Given: `dy/dx = (3x^2 + x)(2x^3 + x^2)^3`

Multiply both sides by `dx`:

`dy = (3x^2 + x)(2x^3 + x^2)^3dx`

Integrate both sides:

`y = int(3x^2 + x)(2x^3 + x^2)^3dx`

`y = int(3x^2 + x)(2x^3 + x^2)^3dx`

Let `u = 2x^3+x^2`, then `du = (6x^2+2x)dx`

This makes the integral become:

`y = 1/2intu^3du`

`y = (1/2)(1/4)u^4+C`

`y = 1/8(2x^3+x^2)^4+ C`