What is the integral of (x+3)(x^2+6x-7) ?

The integral of (x+3)(x^2+6x-7) is (x^4)/4+3x^3+(11x^2)/2-21x+C

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integral of (x+3)(x^2+6x-7)

\int (x + 3) (x^{2} + 6 x - 7) d x

\frac{x ^{4}}{4} + 3 x^{3} + \frac{11 x ^{2}}{2} - 21 x + C

Solution steps

\int (x + 3) (x^{2} + 6 x - 7) d x

Expand

(x + 3) (x^{2} + 6 x - 7) : x^{3} + 9 x^{2} + 11 x - 21

= \int x^{3} + 9 x^{2} + 11 x - 21 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^{3} d x + \int 9 x^{2} d x + \int 11 x d x - \int 21 d x

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

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

\int 11 x d x = \frac{11 x ^{2}}{2}

\int 21 d x = 21 x

= \frac{x ^{4}}{4} + 3 x^{3} + \frac{11 x ^{2}}{2} - 21 x

Add a constant to the solution

= \frac{x ^{4}}{4} + 3 x^{3} + \frac{11 x ^{2}}{2} - 21 x + C

\frac{\partial}{\partial x} (x^{2} + z^{2})

\frac{1}{x ^{2} ( x ^{2} + 4 x + 13 )}

\int_{0}^{2} 3 x d x

\int (3 u + \frac{1}{u}) d u

f (x) = (x - 1)^{2} + 1, at x = 1

What is the integral of (x+3)(x^2+6x-7) ?
The integral of (x+3)(x^2+6x-7) is (x^4)/4+3x^3+(11x^2)/2-21x+C