What is the integral of (3/4 x^2-16x)(3/2 x-16) ?

The integral of (3/4 x^2-16x)(3/2 x-16) is (9x^4)/(32)-12x^3+128x^2+C

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integral of (3/4 x^2-16x)(3/2 x-16)

\int (\frac{3}{4} x^{2} - 16 x) (\frac{3}{2} x - 16) d x

\frac{9 x ^{4}}{32} - 12 x^{3} + 128 x^{2} + C

Solution steps

\int (\frac{3}{4} x^{2} - 16 x) (\frac{3}{2} x - 16) d x

Expand

(\frac{3}{4} x^{2} - 16 x) (\frac{3}{2} x - 16) : \frac{9 x ^{3}}{8} - 36 x^{2} + 256 x

= \int \frac{9 x ^{3}}{8} - 36 x^{2} + 256 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 \frac{9 x ^{3}}{8} d x - \int 36 x^{2} d x + \int 256 x d x

\int \frac{9 x ^{3}}{8} d x = \frac{9 x ^{4}}{32}

\int 36 x^{2} d x = 12 x^{3}

\int 256 x d x = 128 x^{2}

= \frac{9 x ^{4}}{32} - 12 x^{3} + 128 x^{2}

Add a constant to the solution

= \frac{9 x ^{4}}{32} - 12 x^{3} + 128 x^{2} + C

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What is the integral of (3/4 x^2-16x)(3/2 x-16) ?
The integral of (3/4 x^2-16x)(3/2 x-16) is (9x^4)/(32)-12x^3+128x^2+C