What is the integral of (9x^3-5x^2+3x)/x ?

The integral of (9x^3-5x^2+3x)/x is 3x^3-(5x^2)/2+3x+C

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integral of (9x^3-5x^2+3x)/x

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

3 x^{3} - \frac{5 x ^{2}}{2} + 3 x + C

Solution steps

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

Simplify

\frac{9 x ^{3} - 5 x ^{2} + 3 x}{x} : 9 x^{2} - 5 x + 3

= \int 9 x^{2} - 5 x + 3 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 9 x^{2} d x - \int 5 x d x + \int 3 d x

\int 9 x^{2} d x = 3 x^{3}

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 3 d x = 3 x

= 3 x^{3} - \frac{5 x ^{2}}{2} + 3 x

Add a constant to the solution

= 3 x^{3} - \frac{5 x ^{2}}{2} + 3 x + C

What is the integral of (9x^3-5x^2+3x)/x ?
The integral of (9x^3-5x^2+3x)/x is 3x^3-(5x^2)/2+3x+C