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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

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

\int 2 d u = 2 u

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

Substitute back

u = x + 2

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

Simplify

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

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

Add a constant to the solution

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

x y^{^{''}} + 2 y^{^{'}} = 6 x

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

\int \frac{sin ( x )}{cos ^{2} ( x )} d x

x d x - y^{3} d y = 0

\frac{d}{d t} (- A e^{- t})

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