What is the integral of (3x^2+10x)(x^3+5x^2+6) ?

The integral of (3x^2+10x)(x^3+5x^2+6) is (x^6)/2+5x^5+(25x^4)/2+6x^3+30x^2+C

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integral of (3x^2+10x)(x^3+5x^2+6)

\int (3 x^{2} + 10 x) (x^{3} + 5 x^{2} + 6) d x

\frac{x ^{6}}{2} + 5 x^{5} + \frac{25 x ^{4}}{2} + 6 x^{3} + 30 x^{2} + C

Solution steps

\int (3 x^{2} + 10 x) (x^{3} + 5 x^{2} + 6) d x

Expand

(3 x^{2} + 10 x) (x^{3} + 5 x^{2} + 6) : 3 x^{5} + 25 x^{4} + 50 x^{3} + 18 x^{2} + 60 x

= \int 3 x^{5} + 25 x^{4} + 50 x^{3} + 18 x^{2} + 60 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 3 x^{5} d x + \int 25 x^{4} d x + \int 50 x^{3} d x + \int 18 x^{2} d x + \int 60 x d x

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int 25 x^{4} d x = 5 x^{5}

\int 50 x^{3} d x = \frac{25 x ^{4}}{2}

\int 18 x^{2} d x = 6 x^{3}

\int 60 x d x = 30 x^{2}

= \frac{x ^{6}}{2} + 5 x^{5} + \frac{25 x ^{4}}{2} + 6 x^{3} + 30 x^{2}

Add a constant to the solution

= \frac{x ^{6}}{2} + 5 x^{5} + \frac{25 x ^{4}}{2} + 6 x^{3} + 30 x^{2} + C

\frac{d y}{d x} = (y - x + 1)^{2} + 1

x \to 0 lim (\frac{0}{x ^{3}})

y^{^{'}} = 3 - \frac{y}{1000}

y = 49.99 sin (\frac{x}{5.0073}) + 49.99

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

What is the integral of (3x^2+10x)(x^3+5x^2+6) ?
The integral of (3x^2+10x)(x^3+5x^2+6) is (x^6)/2+5x^5+(25x^4)/2+6x^3+30x^2+C