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

The integral of (1-x)x^3(x-2)^2 is -(x^7)/7+(5x^6)/6-(8x^5)/5+x^4+C

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

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

- \frac{x ^{7}}{7} + \frac{5 x ^{6}}{6} - \frac{8 x ^{5}}{5} + x^{4} + C

Solution steps

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

Expand

(1 - x) x^{3} (x - 2)^{2} : - x^{6} + 5 x^{5} - 8 x^{4} + 4 x^{3}

= \int - x^{6} + 5 x^{5} - 8 x^{4} + 4 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 x^{6} d x + \int 5 x^{5} d x - \int 8 x^{4} d x + \int 4 x^{3} d x

\int x^{6} d x = \frac{x ^{7}}{7}

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

\int 8 x^{4} d x = \frac{8 x ^{5}}{5}

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

= - \frac{x ^{7}}{7} + \frac{5 x ^{6}}{6} - \frac{8 x ^{5}}{5} + x^{4}

Add a constant to the solution

= - \frac{x ^{7}}{7} + \frac{5 x ^{6}}{6} - \frac{8 x ^{5}}{5} + x^{4} + C

x y^{^{'}} - 2 y = 2 x^{2} - 2 x - 2

\frac{d}{d x} (5 x^{0.5})

6 x + 9

\int \frac{e ^{x} - 1}{e ^{x} + 1} d x

x \to 1 + lim (x^{2} + 1)

What is the integral of (1-x)x^3(x-2)^2 ?
The integral of (1-x)x^3(x-2)^2 is -(x^7)/7+(5x^6)/6-(8x^5)/5+x^4+C