What is the integral of (x^4)/(x+2) ?

The integral of (x^4)/(x+2) is 12x^2+16x+16ln|x+2|+1/4 (x+2)^4-8/3 (x+2)^3-16+C

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integral of (x^4)/(x+2)

\int \frac{x ^{4}}{x + 2} d x

12 x^{2} + 16 x + 16 ln ∣ x + 2 ∣ + \frac{1}{4} (x + 2)^{4} - \frac{8}{3} (x + 2)^{3} - 16 + C

Solution steps

\int \frac{x ^{4}}{x + 2} d x

Apply u-substitution

: \int \frac{( u - 2 ) ^{4}}{u} d u

= \int \frac{( u - 2 ) ^{4}}{u} d u

Expand

\frac{( u - 2 ) ^{4}}{u} : u^{3} - 8 u^{2} + 24 u - 32 + \frac{16}{u}

= \int u^{3} - 8 u^{2} + 24 u - 32 + \frac{16}{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^{3} d u - \int 8 u^{2} d u + \int 24 u d u - \int 32 d u + \int \frac{16}{u} d u

\int u^{3} d u = \frac{u ^{4}}{4}

\int 8 u^{2} d u = \frac{8 u ^{3}}{3}

\int 24 u d u = 12 u^{2}

\int 32 d u = 32 u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= \frac{u ^{4}}{4} - \frac{8 u ^{3}}{3} + 12 u^{2} - 32 u + 16 ln ∣ u ∣

Substitute back

u = x + 2

= \frac{( x + 2 ) ^{4}}{4} - \frac{8 ( x + 2 ) ^{3}}{3} + 12 (x + 2)^{2} - 32 (x + 2) + 16 ln ∣ x + 2 ∣

Simplify

\frac{( x + 2 ) ^{4}}{4} - \frac{8 ( x + 2 ) ^{3}}{3} + 12 (x + 2)^{2} - 32 (x + 2) + 16 ln ∣ x + 2 ∣ : 12 x^{2} + 16 x + 16 ln ∣ x + 2 ∣ + \frac{1}{4} (x + 2)^{4} - \frac{8}{3} (x + 2)^{3} - 16

= 12 x^{2} + 16 x + 16 ln ∣ x + 2 ∣ + \frac{1}{4} (x + 2)^{4} - \frac{8}{3} (x + 2)^{3} - 16

Add a constant to the solution

= 12 x^{2} + 16 x + 16 ln ∣ x + 2 ∣ + \frac{1}{4} (x + 2)^{4} - \frac{8}{3} (x + 2)^{3} - 16 + C

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What is the integral of (x^4)/(x+2) ?
The integral of (x^4)/(x+2) is 12x^2+16x+16ln|x+2|+1/4 (x+2)^4-8/3 (x+2)^3-16+C