What is the integral of (3x+1)(4-x) ?

The integral of (3x+1)(4-x) is -(3x^3)/2+(x^3)/2+6x^2-(x^2)/2+4x+C

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integral of (3x+1)(4-x)

\int (3 x + 1) (4 - x) d x

- \frac{3 x ^{3}}{2} + \frac{x ^{3}}{2} + 6 x^{2} - \frac{x ^{2}}{2} + 4 x + C

Solution steps

\int (3 x + 1) (4 - x) d x

Apply Integration By Parts

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

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

\int 3 (4 x - \frac{x ^{2}}{2}) d x = 3 (2 x^{2} - \frac{x ^{3}}{6})

= (3 x + 1) (4 x - \frac{x ^{2}}{2}) - 3 (2 x^{2} - \frac{x ^{3}}{6})

Simplify

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

= - \frac{3 x ^{3}}{2} + \frac{x ^{3}}{2} + 6 x^{2} - \frac{x ^{2}}{2} + 4 x

Add a constant to the solution

= - \frac{3 x ^{3}}{2} + \frac{x ^{3}}{2} + 6 x^{2} - \frac{x ^{2}}{2} + 4 x + C

\frac{( x - 2 )}{x + 3}

\frac{d}{d x} (3 x y)

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

\frac{d}{d x} (ln (arctan (8 x^{5})))

\frac{d}{d x} (x^{3} (x - 2)^{2})

What is the integral of (3x+1)(4-x) ?
The integral of (3x+1)(4-x) is -(3x^3)/2+(x^3)/2+6x^2-(x^2)/2+4x+C