What is the integral of (x+2)/(sqrt(x+2)+1) ?

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

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integral of (x+2)/(sqrt(x+2)+1)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{u + 1} d u

= \int \frac{u}{u + 1} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

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

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 w^{2} d w - \int 3 w d w + \int 3 d w - \int \frac{1}{w} d w)

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

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

\int 3 d w = 3 w

\int \frac{1}{w} d w = ln ∣ w ∣

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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