What is the integral of 1/20 (x+1)^2(x-3)(x+4) ?

The integral of 1/20 (x+1)^2(x-3)(x+4) is 1/20 ((x^5)/5+(3x^4)/4-3x^3-(23x^2)/2-12x)+C

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integral of 1/20 (x+1)^2(x-3)(x+4)

\int \frac{1}{20} (x + 1)^{2} (x - 3) (x + 4) d x

\frac{1}{20} (\frac{x ^{5}}{5} + \frac{3 x ^{4}}{4} - 3 x^{3} - \frac{23 x ^{2}}{2} - 12 x) + C

Solution steps

\int \frac{1}{20} (x + 1)^{2} (x - 3) (x + 4) d x

Take the constant out:

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

= \frac{1}{20} \cdot \int (x + 1)^{2} (x - 3) (x + 4) d x

Expand

(x + 1)^{2} (x - 3) (x + 4) : x^{4} + 3 x^{3} - 9 x^{2} - 23 x - 12

= \frac{1}{20} \cdot \int x^{4} + 3 x^{3} - 9 x^{2} - 23 x - 12 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

= \frac{1}{20} (\int x^{4} d x + \int 3 x^{3} d x - \int 9 x^{2} d x - \int 23 x d x - \int 12 d x)

\int x^{4} d x = \frac{x ^{5}}{5}

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

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

\int 23 x d x = \frac{23 x ^{2}}{2}

\int 12 d x = 12 x

= \frac{1}{20} (\frac{x ^{5}}{5} + \frac{3 x ^{4}}{4} - 3 x^{3} - \frac{23 x ^{2}}{2} - 12 x)

Add a constant to the solution

= \frac{1}{20} (\frac{x ^{5}}{5} + \frac{3 x ^{4}}{4} - 3 x^{3} - \frac{23 x ^{2}}{2} - 12 x) + C

y^{^{''}} + 4 y^{^{'}} - 4 y = 0

f (x) = \frac{e ^{x}}{x}

5 cos^{2} (x) + 5 sin (x)

\frac{d}{d x} (\frac{- 6 + x}{6 + x})

\frac{d}{d x} (x (x^{4} + \frac{1}{x ^{2}}))

What is the integral of 1/20 (x+1)^2(x-3)(x+4) ?
The integral of 1/20 (x+1)^2(x-3)(x+4) is 1/20 ((x^5)/5+(3x^4)/4-3x^3-(23x^2)/2-12x)+C