What is the integral of ((4x-5)(2x-7))/3 ?

The integral of ((4x-5)(2x-7))/3 is 1/9 (8x^3-57x^2+105x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((4x-5)(2x-7))/3

\int \frac{( 4 x - 5 ) ( 2 x - 7 )}{3} d x

\frac{1}{9} (8 x^{3} - 57 x^{2} + 105 x) + C

Solution steps

\int \frac{( 4 x - 5 ) ( 2 x - 7 )}{3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int (4 x - 5) (2 x - 7) d x

Apply Integration By Parts

: (2 x - 7) (2 x^{2} - 5 x) - \int 2 (2 x^{2} - 5 x) d x

= \frac{1}{3} ((2 x - 7) (2 x^{2} - 5 x) - \int 2 (2 x^{2} - 5 x) d x)

\int 2 (2 x^{2} - 5 x) d x = \frac{4 x ^{3}}{3} - 5 x^{2}

= \frac{1}{3} ((2 x - 7) (2 x^{2} - 5 x) - (\frac{4 x ^{3}}{3} - 5 x^{2}))

Simplify

\frac{1}{3} ((2 x - 7) (2 x^{2} - 5 x) - (\frac{4 x ^{3}}{3} - 5 x^{2})) : \frac{1}{9} (8 x^{3} - 57 x^{2} + 105 x)

= \frac{1}{9} (8 x^{3} - 57 x^{2} + 105 x)

Add a constant to the solution

= \frac{1}{9} (8 x^{3} - 57 x^{2} + 105 x) + C

\frac{\partial}{\partial x} (ln (x^{2} + 6 y^{2} + 8))

\frac{d y}{d x} = (y + 8)

\int \frac{sec ( θ )}{tan ( θ )} d θ

\frac{d y}{d x} + x \cdot y = - x^{2}

\frac{d}{d x} (x (x - 4)^{2})

What is the integral of ((4x-5)(2x-7))/3 ?
The integral of ((4x-5)(2x-7))/3 is 1/9 (8x^3-57x^2+105x)+C