What is the integral of x^2(x-1)^4 ?

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

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integral of x^2(x-1)^4

\int x^{2} (x - 1)^{4} d x

\frac{x ^{7}}{7} - \frac{2 x ^{6}}{3} + \frac{6 x ^{5}}{5} - x^{4} + \frac{x ^{3}}{3} + C

Solution steps

\int x^{2} (x - 1)^{4} d x

Expand

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

= \int x^{6} - 4 x^{5} + 6 x^{4} - 4 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^{6} d x - \int 4 x^{5} d x + \int 6 x^{4} d x - \int 4 x^{3} d x + \int x^{2} d x

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

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

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

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

= \frac{x ^{7}}{7} - \frac{2 x ^{6}}{3} + \frac{6 x ^{5}}{5} - x^{4} + \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{2 x ^{6}}{3} + \frac{6 x ^{5}}{5} - x^{4} + \frac{x ^{3}}{3} + C

\int 4 x^{3} (4 x^{4} + 5) d x

\int \frac{x}{( 4 - x ^{2} )} d x

\int 5 x^{4} - 3 x^{2} + 4 d x

\int \frac{1}{( x + 1 ) x} d x

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

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