What is the integral of y=12(x-1)^2(x+2) ?

The integral of y=12(x-1)^2(x+2) is 3x^4-18x^2+24x+C

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integral of y=12(x-1)^2(x+2)

integral

y = 12 (x - 1)^{2} (x + 2)

3 x^{4} - 18 x^{2} + 24 x + C

Solution steps

\int 12 (x - 1)^{2} (x + 2) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \int (x - 1)^{2} (x + 2) d x

Apply Integration By Parts

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

= 12 ((x + 2) (\frac{x ^{3}}{3} - x^{2} + x) - \int \frac{x ^{3}}{3} - x^{2} + x d x)

\int \frac{x ^{3}}{3} - x^{2} + x d x = \frac{x ^{4}}{12} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2}

= 12 ((x + 2) (\frac{x ^{3}}{3} - x^{2} + x) - (\frac{x ^{4}}{12} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2}))

Simplify

12 ((x + 2) (\frac{x ^{3}}{3} - x^{2} + x) - (\frac{x ^{4}}{12} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2})) : 3 x^{4} - 18 x^{2} + 24 x

= 3 x^{4} - 18 x^{2} + 24 x

Add a constant to the solution

= 3 x^{4} - 18 x^{2} + 24 x + C

lo g_{10} (x^{2} + 1)

(1 + x^{2})^{0.5}

\frac{tan ^{2} ( x )}{cos ( x )}

\frac{( 4 - x )}{x}

\frac{( e ^{- t} - 4 )}{e ^{- t}}

What is the integral of y=12(x-1)^2(x+2) ?
The integral of y=12(x-1)^2(x+2) is 3x^4-18x^2+24x+C