What is the integral of (8x+3)(7-x) ?

The integral of (8x+3)(7-x) is -4x^3+25x^2+21x+(4x^3)/3+(3x^2)/2+C

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integral of (8x+3)(7-x)

\int (8 x + 3) (7 - x) d x

- 4 x^{3} + 25 x^{2} + 21 x + \frac{4 x ^{3}}{3} + \frac{3 x ^{2}}{2} + C

Solution steps

\int (8 x + 3) (7 - x) d x

Apply Integration By Parts

: (7 - x) (4 x^{2} + 3 x) - \int - 4 x^{2} - 3 x d x

= (7 - x) (4 x^{2} + 3 x) - \int - 4 x^{2} - 3 x d x

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

= (7 - x) (4 x^{2} + 3 x) - (- \frac{4 x ^{3}}{3} - \frac{3 x ^{2}}{2})

Simplify

(7 - x) (4 x^{2} + 3 x) - (- \frac{4 x ^{3}}{3} - \frac{3 x ^{2}}{2}) : - 4 x^{3} + 25 x^{2} + 21 x + \frac{4 x ^{3}}{3} + \frac{3 x ^{2}}{2}

= - 4 x^{3} + 25 x^{2} + 21 x + \frac{4 x ^{3}}{3} + \frac{3 x ^{2}}{2}

Add a constant to the solution

= - 4 x^{3} + 25 x^{2} + 21 x + \frac{4 x ^{3}}{3} + \frac{3 x ^{2}}{2} + C

x \to - \infty lim (- 3 x + x)

h = (t + 1)^{\frac{2}{3}} (2 t^{2} - 1)^{3}

\int \frac{x x}{2} d x

f (x) = \frac{( 4 x + x ^{2} )}{( 2 + x ) ^{2}}

\int \frac{14 - 14 x}{1 - x} d x

What is the integral of (8x+3)(7-x) ?
The integral of (8x+3)(7-x) is -4x^3+25x^2+21x+(4x^3)/3+(3x^2)/2+C