What is the integral from 0 to 1 of (x^3)/(x^2+6x+9) ?

The integral from 0 to 1 of (x^3)/(x^2+6x+9) is 27(2ln(2)-ln(3))-31/4

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integral from 0 to 1 of (x^3)/(x^2+6x+9)

\int_{0}^{1} \frac{x ^{3}}{x ^{2} + 6 x + 9} d x

27 (2 ln (2) - ln (3)) - \frac{31}{4}

Decimal Notation

0.01741 \dots

Solution steps

\int_{0}^{1} \frac{x ^{3}}{x ^{2} + 6 x + 9} d x

Complete the square

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

= \int_{0}^{1} \frac{x ^{3}}{( x + 3 ) ^{2}} d x

Apply u-substitution

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

= \int_{3}^{4} \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_{3}^{4} 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_{3}^{4} u d u - \int_{3}^{4} 9 d u + \int_{3}^{4} \frac{27}{u} d u - \int_{3}^{4} \frac{27}{u ^{2}} d u

\int_{3}^{4} u d u = \frac{7}{2}

\int_{3}^{4} 9 d u = 9

\int_{3}^{4} \frac{27}{u} d u = 27 (2 ln (2) - ln (3))

\int_{3}^{4} \frac{27}{u ^{2}} d u = \frac{9}{4}

= \frac{7}{2} - 9 + 27 (2 ln (2) - ln (3)) - \frac{9}{4}

Combine the fractions

- 9 + \frac{7}{2} - \frac{9}{4} : - \frac{31}{4}

= 27 (2 ln (2) - ln (3)) - \frac{31}{4}

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What is the integral from 0 to 1 of (x^3)/(x^2+6x+9) ?
The integral from 0 to 1 of (x^3)/(x^2+6x+9) is 27(2ln(2)-ln(3))-31/4