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

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

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

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

\frac{x ^{5}}{5} - \frac{x ^{4}}{2} + \frac{x ^{3}}{3} + C

Solution steps

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

Expand

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

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

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

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

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

= \frac{x ^{5}}{5} - \frac{x ^{4}}{2} + \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{x ^{4}}{2} + \frac{x ^{3}}{3} + C

t a y l or \frac{2}{( 3 x + 1 )}, x = 2

\int - x e^{- l n (x)} d x

t an g e n t f (x) = 2 x^{3} + 3 x^{2}, at x = 1

\frac{d}{d x} (25 (x^{2} - y^{2}))

\frac{d}{d x} (\frac{1 0 ^{x}}{1 0 ^{x} + 4})

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