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

The integral of (x^2)/(sqrt(x+2)) is 2(1/5 (x+2)^{5/2}-4/3 (x+2)^{3/2}+4sqrt(x+2))+C

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integral of (x^2)/(sqrt(x+2))

\int \frac{x ^{2}}{x + 2} d x

2 (\frac{1}{5} (x + 2)^{\frac{5}{2}} - \frac{4}{3} (x + 2)^{\frac{3}{2}} + 4 x + 2) + C

Solution steps

\int \frac{x ^{2}}{x + 2} d x

Apply u-substitution

: \int 2 (u^{2} - 2)^{2} d u

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

Take the constant out:

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

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

Expand

(u^{2} - 2)^{2} : u^{4} - 4 u^{2} + 4

= 2 \cdot \int u^{4} - 4 u^{2} + 4 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^{4} d u - \int 4 u^{2} d u + \int 4 d u)

\int u^{4} d u = \frac{u ^{5}}{5}

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

\int 4 d u = 4 u

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

Substitute back

u = x + 2

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

Simplify

2 (\frac{( x + 2 ) ^{5}}{5} - \frac{4 ( x + 2 ) ^{3}}{3} + 4 x + 2) : 2 (\frac{1}{5} (x + 2)^{\frac{5}{2}} - \frac{4}{3} (x + 2)^{\frac{3}{2}} + 4 x + 2)

= 2 (\frac{1}{5} (x + 2)^{\frac{5}{2}} - \frac{4}{3} (x + 2)^{\frac{3}{2}} + 4 x + 2)

Add a constant to the solution

= 2 (\frac{1}{5} (x + 2)^{\frac{5}{2}} - \frac{4}{3} (x + 2)^{\frac{3}{2}} + 4 x + 2) + C

y = 5 x^{3} + 4 x^{2} - 6, (1, 3)

\int e^{- 2 x} \cdot sin (x) d x

x \to 2 lim ((- 2)^{3})

\frac{\partial}{\partial x} (\frac{x y}{( x + y )})

\frac{d}{d x} (- \frac{1}{1 + x ^{2}})

What is the integral of (x^2)/(sqrt(x+2)) ?
The integral of (x^2)/(sqrt(x+2)) is 2(1/5 (x+2)^{5/2}-4/3 (x+2)^{3/2}+4sqrt(x+2))+C