How do you integrate $\frac{x^{3} + 2 x^{2} - x}{x} d x$ $(x^3+2x^2-x) / x dx$ ?

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How do you integrate $\frac{x^{3} + 2 x^{2} - x}{x} d x$ $(x^3+2x^2-x) / x dx$ ?

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Answer 1 · 2018-07-01T16:24:42

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

Explanation:

At first we simplify the integrand

$\frac{x^{3} + 2 x^{2} - x}{x} = \frac{x^{3}}{x} + 2 \frac{x^{2}}{x} - \frac{x}{x} = x^{2} + 2 x - 1$ $(x^3+2x^2-x)/x=x^3/x+2x^2/x-x/x=x^2+2x-1$
Now we can integrate the result:

$\int (x^{2} + 2 x - 1) d x = \frac{x^{3}}{3} + 2 \frac{x^{2}}{2} - x + C = \frac{x^{3}}{3} + x^{2} - x + C$ $int(x^2+2x-1)dx=x^3/3+2x^2/2-x+C=x^3/3+x^2-x+C$

we have used that

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $int x^ndx=x^(n+1)/(n+1)+C$ if $n \neq - 1$ $n ne -1$

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How do you integrate $\frac{x^{3} + 2 x^{2} - x}{x} d x$ $(x^3+2x^2-x) / x dx$ ?

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Explanation:

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How do you integrate $\frac{x^{3} + 2 x^{2} - x}{x} d x$ $(x^3+2x^2-x) / x dx$ ?