What is the integral of zsqrt(z-8) ?

The integral of zsqrt(z-8) is 2/5 (z-8)^{5/2}+16/3 (z-8)^{3/2}+C

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integral of zsqrt(z-8)

\int z z - 8 d z

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

Solution steps

\int z z - 8 d z

Apply u-substitution

= \int (u + 8) u d u

Expand

(u + 8) u : u^{\frac{3}{2}} + 8 u

= \int u^{\frac{3}{2}} + 8 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 8 u d u

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

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

= \frac{2}{5} u^{\frac{5}{2}} + \frac{16}{3} u^{\frac{3}{2}}

Substitute back

u = z - 8

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

Add a constant to the solution

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

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