What is the integral of (x^3)/(x^2+6x+9) ?

The integral of (x^3)/(x^2+6x+9) is 1/2 (x+3)^2-9(x+3)+27ln|x+3|+(27)/(x+3)+C

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integral of (x^3)/(x^2+6x+9)

\int \frac{x ^{3}}{x ^{2} + 6 x + 9} d x

\frac{1}{2} (x + 3)^{2} - 9 (x + 3) + 27 ln ∣ x + 3 ∣ + \frac{27}{x + 3} + C

Solution steps

\int \frac{x ^{3}}{x ^{2} + 6 x + 9} d x

Complete the square

x^{2} + 6 x + 9 : (x + 3)^{2}

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

Apply u-substitution

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

Expand

\frac{( u - 3 ) ^{3}}{u ^{2}} : u - 9 + \frac{27}{u} - \frac{27}{u ^{2}}

= \int u - 9 + \frac{27}{u} - \frac{27}{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 9 d u + \int \frac{27}{u} d u - \int \frac{27}{u ^{2}} d u

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

\int 9 d u = 9 u

\int \frac{27}{u} d u = 27 ln ∣ u ∣

\int \frac{27}{u ^{2}} d u = - \frac{27}{u}

= \frac{u ^{2}}{2} - 9 u + 27 ln ∣ u ∣ - (- \frac{27}{u})

Substitute back

u = x + 3

= \frac{( x + 3 ) ^{2}}{2} - 9 (x + 3) + 27 ln ∣ x + 3 ∣ - (- \frac{27}{x + 3})

Simplify

\frac{( x + 3 ) ^{2}}{2} - 9 (x + 3) + 27 ln ∣ x + 3 ∣ - (- \frac{27}{x + 3}) : \frac{1}{2} (x + 3)^{2} - 9 (x + 3) + 27 ln ∣ x + 3 ∣ + \frac{27}{x + 3}

= \frac{1}{2} (x + 3)^{2} - 9 (x + 3) + 27 ln ∣ x + 3 ∣ + \frac{27}{x + 3}

Add a constant to the solution

= \frac{1}{2} (x + 3)^{2} - 9 (x + 3) + 27 ln ∣ x + 3 ∣ + \frac{27}{x + 3} + C

What is the integral of (x^3)/(x^2+6x+9) ?
The integral of (x^3)/(x^2+6x+9) is 1/2 (x+3)^2-9(x+3)+27ln|x+3|+(27)/(x+3)+C