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

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

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

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

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

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Add a constant to the solution

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

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