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

The integral of x/(2+sqrt(x)) is 2(-3x+1/3 (sqrt(x)+2)^3-8ln|sqrt(x)+2|+12)+C

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

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

2 (- 3 x + \frac{1}{3} (x + 2)^{3} - 8 ln x + 2 + 12) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{3}}{2 + u} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{( v - 2 ) ^{3}}{v} d v

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

Expand

\frac{( v - 2 ) ^{3}}{v} : v^{2} - 6 v + 12 - \frac{8}{v}

= 2 \cdot \int v^{2} - 6 v + 12 - \frac{8}{v} d v

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 v^{2} d v - \int 6 v d v + \int 12 d v - \int \frac{8}{v} d v)

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

\int 6 v d v = 3 v^{2}

\int 12 d v = 12 v

\int \frac{8}{v} d v = 8 ln ∣ v ∣

= 2 (\frac{v ^{3}}{3} - 3 v^{2} + 12 v - 8 ln ∣ v ∣)

Substitute back

= 2 (\frac{( 2 + x ) ^{3}}{3} - 3 (2 + x)^{2} + 12 (2 + x) - 8 ln 2 + x)

Simplify

2 (\frac{( 2 + x ) ^{3}}{3} - 3 (2 + x)^{2} + 12 (2 + x) - 8 ln 2 + x) : 2 (- 3 x + \frac{1}{3} (x + 2)^{3} - 8 ln x + 2 + 12)

= 2 (- 3 x + \frac{1}{3} (x + 2)^{3} - 8 ln x + 2 + 12)

Add a constant to the solution

= 2 (- 3 x + \frac{1}{3} (x + 2)^{3} - 8 ln x + 2 + 12) + C

f (x) = (- x^{3} - x^{2} + x) cos (x)

\frac{d y}{d x} = x + \frac{1}{5} y

\frac{d}{d x} (\frac{x ^{2} - 4}{x + 1})

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + y = 0

f (x) = x^{8} - 46

What is the integral of x/(2+sqrt(x)) ?
The integral of x/(2+sqrt(x)) is 2(-3x+1/3 (sqrt(x)+2)^3-8ln|sqrt(x)+2|+12)+C