What is the integral of (2x-5)(3x^2+4x) ?

The integral of (2x-5)(3x^2+4x) is (3x^4)/2-(7x^3)/3-10x^2+C

Popular >

integral of (2x-5)(3x^2+4x)

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

\frac{3 x ^{4}}{2} - \frac{7 x ^{3}}{3} - 10 x^{2} + C

Solution steps

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

Expand

(2 x - 5) (3 x^{2} + 4 x) : 6 x^{3} - 7 x^{2} - 20 x

= \int 6 x^{3} - 7 x^{2} - 20 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 6 x^{3} d x - \int 7 x^{2} d x - \int 20 x d x

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

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

\int 20 x d x = 10 x^{2}

= \frac{3 x ^{4}}{2} - \frac{7 x ^{3}}{3} - 10 x^{2}

Add a constant to the solution

= \frac{3 x ^{4}}{2} - \frac{7 x ^{3}}{3} - 10 x^{2} + C

\int \frac{5 x - 6}{( 2 x + 3 ) ^{9}} d x

\int 6 x (6 x^{2} + 1)^{\frac{5}{2}} d x

\int (x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{7} d x

\int \frac{sec ^{2} ( x )}{tan ( x ) + 1} d x

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

What is the integral of (2x-5)(3x^2+4x) ?
The integral of (2x-5)(3x^2+4x) is (3x^4)/2-(7x^3)/3-10x^2+C