What is the integral of (x^3)/((x+2)) ?

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

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integral of (x^3)/((x+2))

integral

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

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

Solution steps

\int \frac{x ^{3}}{x + 2} d x

Apply u-substitution

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

Expand

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

= \int u^{2} - 6 u + 12 - \frac{8}{u} 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

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

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

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

\int 12 d u = 12 u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

= \frac{u ^{3}}{3} - 3 u^{2} + 12 u - 8 ln ∣ u ∣

Substitute back

u = x + 2

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

Simplify

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

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

Add a constant to the solution

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

in t e g r a l \frac{1}{( a ^{2} - x ^{2} )}

in t e g r a l \frac{x ^{2}}{( 1 - x )}

in t e g r a l \frac{k \cdot x}{( 1 + x ^{2} )}

in t e g r a l lo g_{10} (x + 2)

in t e g r a l \frac{x}{( 1 + x ^{2} )}

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