What is the integral of x/(1+sqrt(x)) ?

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

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integral of x/(1+sqrt(x))

\int \frac{x}{1 + x} d x

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

Solution steps

\int \frac{x}{1 + x} d x

Apply u-substitution

: \int \frac{2 u ^{3}}{1 + u} d u

= \int \frac{2 u ^{3}}{1 + 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}}{1 + u} d u

Apply u-substitution

: \int \frac{( v - 1 ) ^{3}}{v} d v

= 2 \cdot \int \frac{( v - 1 ) ^{3}}{v} d v

Expand

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

= 2 \cdot \int v^{2} - 3 v + 3 - \frac{1}{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 3 v d v + \int 3 d v - \int \frac{1}{v} d v)

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

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

\int 3 d v = 3 v

\int \frac{1}{v} d v = ln ∣ v ∣

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (- 2 x)

\frac{\partial}{\partial x} (x e^{- 2 x^{2} - 2 y^{2}})

\int 2 x^{2} + 1 d x

\frac{3 x ^{2} + 3 x - 2}{2 x x}

(5.5) (2.3)

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