What is the integral of zsqrt(z-1) ?

The integral of zsqrt(z-1) is 2/5 (z-1)^{5/2}+2/3 (z-1)^{3/2}+C

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integral of zsqrt(z-1)

\int z z - 1 d z

\frac{2}{5} (z - 1)^{\frac{5}{2}} + \frac{2}{3} (z - 1)^{\frac{3}{2}} + C

Solution steps

\int z z - 1 d z

Apply u-substitution

= \int (u + 1) u d u

Expand

(u + 1) u : u^{\frac{3}{2}} + u

= \int u^{\frac{3}{2}} + u d u

Apply the Sum Rule:

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

= \int u^{\frac{3}{2}} d u + \int u d u

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = z - 1

= \frac{2}{5} (z - 1)^{\frac{5}{2}} + \frac{2}{3} (z - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (z - 1)^{\frac{5}{2}} + \frac{2}{3} (z - 1)^{\frac{3}{2}} + C

What is the integral of zsqrt(z-1) ?
The integral of zsqrt(z-1) is 2/5 (z-1)^{5/2}+2/3 (z-1)^{3/2}+C