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

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

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

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

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

Solution steps

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

Simplify

\frac{x ^{3} - 2 x ^{2} + 4 x}{x} : x^{2} - 2 x + 4

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

\frac{d y}{d x} = y^{2} - 9

\int \frac{ln ( t )}{3} d t

\int \frac{x}{( x + 3 ) ^{5}} d x

\int (x + 1) sin (2 x - 3) d x

y^{^{''}} - 20 y^{^{'}} + 181 y = 0

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