What is the integral of xsqrt(x-8) ?

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

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integral of xsqrt(x-8)

\int x x - 8 d x

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

Solution steps

\int x x - 8 d x

Apply u-substitution

: \int (u + 8) u d u

= \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 = x - 8

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

Add a constant to the solution

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

3 x^{2} + x y + 3 y^{2} = 7, (1, 1)

\frac{sin ( t )}{2 + 2 cos ( t )}

\frac{1}{( s + 1 ) ^{2} + 4}

\int \frac{x}{x + 3} d x

\frac{d}{d x} (\frac{d y}{d x}) - y = 0

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