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

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

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

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

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

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

Add a constant to the solution

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

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