What is the integral of x(x+4)(2x-1) ?

The integral of x(x+4)(2x-1) is (x^4)/2+(7x^3)/3-2x^2+C

integral of x(x+4)(2x-1)

\int x (x + 4) (2 x - 1) d x

\frac{x ^{4}}{2} + \frac{7 x ^{3}}{3} - 2 x^{2} + C

Solution steps

\int x (x + 4) (2 x - 1) d x

Expand

x (x + 4) (2 x - 1) : 2 x^{3} + 7 x^{2} - 4 x

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

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

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 4 x d x = 2 x^{2}

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

Add a constant to the solution

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

\int \frac{1}{( x ln ( x ) )} d x

in t e g r a l sin (3 x)

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\int (\frac{x ^{2}}{x - 1}) d x

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