What is the integral of (x^3)/(x^2+8x+16) ?

The integral of (x^3)/(x^2+8x+16) is 1/2 (x+4)^2-12(x+4)+48ln|x+4|+(64)/(x+4)+C

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integral of (x^3)/(x^2+8x+16)

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

\frac{1}{2} (x + 4)^{2} - 12 (x + 4) + 48 ln ∣ x + 4 ∣ + \frac{64}{x + 4} + C

Solution steps

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

Complete the square

x^{2} + 8 x + 16 : (x + 4)^{2}

= \int \frac{x ^{3}}{( x + 4 ) ^{2}} d x

Apply u-substitution

: \int \frac{( u - 4 ) ^{3}}{u ^{2}} d u

= \int \frac{( u - 4 ) ^{3}}{u ^{2}} d u

Expand

\frac{( u - 4 ) ^{3}}{u ^{2}} : u - 12 + \frac{48}{u} - \frac{64}{u ^{2}}

= \int u - 12 + \frac{48}{u} - \frac{64}{u ^{2}} 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 d u - \int 12 d u + \int \frac{48}{u} d u - \int \frac{64}{u ^{2}} d u

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

\int 12 d u = 12 u

\int \frac{48}{u} d u = 48 ln ∣ u ∣

\int \frac{64}{u ^{2}} d u = - \frac{64}{u}

= \frac{u ^{2}}{2} - 12 u + 48 ln ∣ u ∣ - (- \frac{64}{u})

Substitute back

u = x + 4

= \frac{( x + 4 ) ^{2}}{2} - 12 (x + 4) + 48 ln ∣ x + 4 ∣ - (- \frac{64}{x + 4})

Simplify

\frac{( x + 4 ) ^{2}}{2} - 12 (x + 4) + 48 ln ∣ x + 4 ∣ - (- \frac{64}{x + 4}) : \frac{1}{2} (x + 4)^{2} - 12 (x + 4) + 48 ln ∣ x + 4 ∣ + \frac{64}{x + 4}

= \frac{1}{2} (x + 4)^{2} - 12 (x + 4) + 48 ln ∣ x + 4 ∣ + \frac{64}{x + 4}

Add a constant to the solution

= \frac{1}{2} (x + 4)^{2} - 12 (x + 4) + 48 ln ∣ x + 4 ∣ + \frac{64}{x + 4} + C

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What is the integral of (x^3)/(x^2+8x+16) ?
The integral of (x^3)/(x^2+8x+16) is 1/2 (x+4)^2-12(x+4)+48ln|x+4|+(64)/(x+4)+C