What is the integral of (x^2(x^3-2x+1)) ?

The integral of (x^2(x^3-2x+1)) is (x^6)/6-(x^4)/2+(x^3)/3+C

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integral of (x^2(x^3-2x+1))

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

\frac{x ^{6}}{6} - \frac{x ^{4}}{2} + \frac{x ^{3}}{3} + C

Solution steps

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

Expand

x^{2} (x^{3} - 2 x + 1) : x^{5} - 2 x^{3} + x^{2}

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

\int x^{5} d x = \frac{x ^{6}}{6}

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

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

= \frac{x ^{6}}{6} - \frac{x ^{4}}{2} + \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{6}}{6} - \frac{x ^{4}}{2} + \frac{x ^{3}}{3} + C

\frac{d}{d x} (\frac{8 x ^{2} + 8 x + 6}{x})

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

\int \frac{29 x ^{4} + 29 x ^{3} + 2 x + 1}{x ^{2} + x} d x

\frac{d}{d x} y = (sin (\frac{x}{2}) - cos (\frac{x}{2}))^{2}

\int \frac{5 - x}{2 x ^{2} + x - 1} d x

What is the integral of (x^2(x^3-2x+1)) ?
The integral of (x^2(x^3-2x+1)) is (x^6)/6-(x^4)/2+(x^3)/3+C