What is the integral of (3x-h)^3 ?

The integral of (3x-h)^3 is (27x^4)/4-9hx^3+(9h^2x^2)/2-h^3x+C

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integral of (3x-h)^3

\int (3 x - h)^{3} d x

\frac{27 x ^{4}}{4} - 9 h x^{3} + \frac{9 h ^{2} x ^{2}}{2} - h^{3} x + C

Solution steps

\int (3 x - h)^{3} d x

Expand

(3 x - h)^{3} : 27 x^{3} - 27 h x^{2} + 9 h^{2} x - h^{3}

= \int 27 x^{3} - 27 h x^{2} + 9 h^{2} x - h^{3} 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 27 x^{3} d x - \int 27 h x^{2} d x + \int 9 h^{2} x d x - \int h^{3} d x

\int 27 x^{3} d x = \frac{27 x ^{4}}{4}

\int 27 h x^{2} d x = 9 h x^{3}

\int 9 h^{2} x d x = \frac{9 h ^{2} x ^{2}}{2}

\int h^{3} d x = h^{3} x

= \frac{27 x ^{4}}{4} - 9 h x^{3} + \frac{9 h ^{2} x ^{2}}{2} - h^{3} x

Add a constant to the solution

= \frac{27 x ^{4}}{4} - 9 h x^{3} + \frac{9 h ^{2} x ^{2}}{2} - h^{3} x + C

What is the integral of (3x-h)^3 ?
The integral of (3x-h)^3 is (27x^4)/4-9hx^3+(9h^2x^2)/2-h^3x+C