What is the integral of 2xsqrt(x+2) ?

The integral of 2xsqrt(x+2) is 2(2/5 (x+2)^{5/2}-4/3 (x+2)^{3/2})+C

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integral of 2xsqrt(x+2)

\int 2 x x + 2 d x

2 (\frac{2}{5} (x + 2)^{\frac{5}{2}} - \frac{4}{3} (x + 2)^{\frac{3}{2}}) + C

Solution steps

\int 2 x x + 2 d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int x x + 2 d x

Apply u-substitution

: \int (u - 2) u d u

= 2 \cdot \int (u - 2) u d u

Expand

(u - 2) u : u^{\frac{3}{2}} - 2 u

= 2 \cdot \int u^{\frac{3}{2}} - 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

= 2 (\int u^{\frac{3}{2}} d u - \int 2 u d u)

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

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

= 2 (\frac{2}{5} u^{\frac{5}{2}} - \frac{4}{3} u^{\frac{3}{2}})

Substitute back

u = x + 2

= 2 (\frac{2}{5} (x + 2)^{\frac{5}{2}} - \frac{4}{3} (x + 2)^{\frac{3}{2}})

Add a constant to the solution

= 2 (\frac{2}{5} (x + 2)^{\frac{5}{2}} - \frac{4}{3} (x + 2)^{\frac{3}{2}}) + C

y^{^{''}} - 22 y^{^{'}} + 121 y = 0

n \to \infty lim (i = 1 \sum n \frac{1}{2 n + i})

\frac{\partial}{\partial x} (5 x^{2} y^{3} + 2 x^{2} + 7 y)

f (x) = e^{- x}, 5

y = ln (5 x \cdot (2 x^{3} - 4 x)^{5})

What is the integral of 2xsqrt(x+2) ?
The integral of 2xsqrt(x+2) is 2(2/5 (x+2)^{5/2}-4/3 (x+2)^{3/2})+C