What is the integral of sqrt(z)(4(z-1)^2+1/(z^3)) ?

The integral of sqrt(z)(4(z-1)^2+1/(z^3)) is -2/(3z^{3/2)}+8/7 z^{7/2}-16/5 z^{5/2}+8/3 z^{3/2}+C

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integral of sqrt(z)(4(z-1)^2+1/(z^3))

\int z (4 (z - 1)^{2} + \frac{1}{z ^{3}}) d z

- \frac{2}{3 z ^{\frac{3}{2}}} + \frac{8}{7} z^{\frac{7}{2}} - \frac{16}{5} z^{\frac{5}{2}} + \frac{8}{3} z^{\frac{3}{2}} + C

Solution steps

\int z (4 (z - 1)^{2} + \frac{1}{z ^{3}}) d z

Expand

z (4 (z - 1)^{2} + \frac{1}{z ^{3}}) : \frac{1}{z ^{\frac{5}{2}}} + 4 z^{\frac{5}{2}} - 8 z^{\frac{3}{2}} + 4 z^{\frac{1}{2}}

= \int \frac{1}{z ^{\frac{5}{2}}} + 4 z^{\frac{5}{2}} - 8 z^{\frac{3}{2}} + 4 z^{\frac{1}{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 \frac{1}{z ^{\frac{5}{2}}} d z + \int 4 z^{\frac{5}{2}} d z - \int 8 z^{\frac{3}{2}} d z + \int 4 z^{\frac{1}{2}} d z

\int \frac{1}{z ^{\frac{5}{2}}} d z = - \frac{2}{3 z ^{\frac{3}{2}}}

\int 4 z^{\frac{5}{2}} d z = \frac{8}{7} z^{\frac{7}{2}}

\int 8 z^{\frac{3}{2}} d z = \frac{16}{5} z^{\frac{5}{2}}

\int 4 z^{\frac{1}{2}} d z = \frac{8}{3} z^{\frac{3}{2}}

= - \frac{2}{3 z ^{\frac{3}{2}}} + \frac{8}{7} z^{\frac{7}{2}} - \frac{16}{5} z^{\frac{5}{2}} + \frac{8}{3} z^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3 z ^{\frac{3}{2}}} + \frac{8}{7} z^{\frac{7}{2}} - \frac{16}{5} z^{\frac{5}{2}} + \frac{8}{3} z^{\frac{3}{2}} + C

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What is the integral of sqrt(z)(4(z-1)^2+1/(z^3)) ?
The integral of sqrt(z)(4(z-1)^2+1/(z^3)) is -2/(3z^{3/2)}+8/7 z^{7/2}-16/5 z^{5/2}+8/3 z^{3/2}+C