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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Simplify

2 (\frac{( x + 1 ) ^{3}}{3} - \frac{3 ( x + 1 ) ^{2}}{2} + 3 (x + 1) - ln x + 1) : - 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{4}{S} + \frac{6}{S ^{5}} - \frac{1}{S + 8}

\frac{d}{d x} (- \frac{2}{5} x^{\frac{5}{2}} + 32 x^{\frac{1}{2}})

\frac{d}{d x} (x x + 8)

\int x^{3} - 8 x^{2} + 16 x + 1 d x

3 x^{\frac{1}{3}} - 5 x^{\frac{4}{3}}

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