What is the integral of ((3x^2)/2+4x)(4x+2/3) ?

The integral of ((3x^2)/2+4x)(4x+2/3) is (3x^4)/2+(17x^3)/3+(4x^2)/3+C

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integral of ((3x^2)/2+4x)(4x+2/3)

\int (\frac{3 x ^{2}}{2} + 4 x) (4 x + \frac{2}{3}) d x

\frac{3 x ^{4}}{2} + \frac{17 x ^{3}}{3} + \frac{4 x ^{2}}{3} + C

Solution steps

\int (\frac{3 x ^{2}}{2} + 4 x) (4 x + \frac{2}{3}) d x

Simplify

(\frac{3 x ^{2}}{2} + 4 x) (4 x + \frac{2}{3}) : 6 x^{3} + 17 x^{2} + \frac{8}{3} x

= \int 6 x^{3} + 17 x^{2} + \frac{8}{3} x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 6 x^{3} d x + \int 17 x^{2} d x + \int \frac{8}{3} x d x

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

\int 17 x^{2} d x = \frac{17 x ^{3}}{3}

\int \frac{8}{3} x d x = \frac{4 x ^{2}}{3}

= \frac{3 x ^{4}}{2} + \frac{17 x ^{3}}{3} + \frac{4 x ^{2}}{3}

Add a constant to the solution

= \frac{3 x ^{4}}{2} + \frac{17 x ^{3}}{3} + \frac{4 x ^{2}}{3} + C

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What is the integral of ((3x^2)/2+4x)(4x+2/3) ?
The integral of ((3x^2)/2+4x)(4x+2/3) is (3x^4)/2+(17x^3)/3+(4x^2)/3+C