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

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

integral of ((3x-4)4x^3)/(x^2)

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

4 (x^{3} - 2 x^{2}) + C

Solution steps

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

Take the constant out:

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

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

Simplify

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

= 4 \cdot \int x (3 x - 4) d x

Expand

x (3 x - 4) : 3 x^{2} - 4 x

= 4 \cdot \int 3 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

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

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

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

= 4 (x^{3} - 2 x^{2})

Add a constant to the solution

= 4 (x^{3} - 2 x^{2}) + C

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

\int \frac{1}{9 - ( x + 2 ) ^{2}} d x

\int sin (m x) cos (n x) d x

\int \frac{2 x ^{3} - x + 3}{x + 1} d x

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

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