What is the integral of (x+3)(x+1)x(x-2) ?

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

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integral of (x+3)(x+1)x(x-2)

\int (x + 3) (x + 1) x (x - 2) d x

\frac{x ^{5}}{5} + \frac{x ^{4}}{2} - \frac{5 x ^{3}}{3} - 3 x^{2} + C

Solution steps

\int (x + 3) (x + 1) x (x - 2) d x

Expand

(x + 3) (x + 1) x (x - 2) : x^{4} + 2 x^{3} - 5 x^{2} - 6 x

= \int x^{4} + 2 x^{3} - 5 x^{2} - 6 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 x^{4} d x + \int 2 x^{3} d x - \int 5 x^{2} d x - \int 6 x d x

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

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

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

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

= \frac{x ^{5}}{5} + \frac{x ^{4}}{2} - \frac{5 x ^{3}}{3} - 3 x^{2}

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{x ^{4}}{2} - \frac{5 x ^{3}}{3} - 3 x^{2} + C

y^{^{''}} - 6 y^{^{'}} - 4 y = 0

\int sin (7 y) d y

\int_{a}^{\infty} f (x) d x

y^{^{''}} - y^{^{'}} - 6 y = 6 e^{3 x} + 2 e^{- 2 x}

x^{4} + 5 e^{x}

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