What is the integral of (x+2)sqrt(x-2) ?

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

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integral of (x+2)sqrt(x-2)

\int (x + 2) x - 2 d x

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

Solution steps

\int (x + 2) x - 2 d x

Apply u-substitution

: \int u (u + 4) d u

= \int u (u + 4) d u

Expand

u (u + 4) : u^{\frac{3}{2}} + 4 u

= \int u^{\frac{3}{2}} + 4 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 4 u d u

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

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

= \frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}}

Substitute back

u = x - 2

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

Add a constant to the solution

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

\int \frac{1}{( x ^{4} - 1 )} d x

\int_{2}^{\frac{1}{x}} tan^{2} (t) d t

f (x) = x \cdot cot (x), at x = \frac{π}{4}

f (x) = \frac{x + x}{x ^{3}}

\frac{d}{d x} (\frac{y}{y + x ^{2}})

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