What is the integral from 0 to 2 of (x^3)/(x^2+4x+4) ?

The integral from 0 to 2 of (x^3)/(x^2+4x+4) is 12ln(2)-8

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integral from 0 to 2 of (x^3)/(x^2+4x+4)

\int_{0}^{2} \frac{x ^{3}}{x ^{2} + 4 x + 4} d x

12 ln (2) - 8

Decimal Notation

0.31776 \dots

Solution steps

\int_{0}^{2} \frac{x ^{3}}{x ^{2} + 4 x + 4} d x

Apply Integration By Parts

: [- \frac{x ^{3}}{x + 2} - \int - \frac{3 x ^{2}}{x + 2} d x]_{0}^{2}

= [- \frac{x ^{3}}{x + 2} - \int - \frac{3 x ^{2}}{x + 2} d x]_{0}^{2}

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

= [- \frac{x ^{3}}{x + 2} - (- 3 (\frac{( x + 2 ) ^{2}}{2} - 4 (x + 2) + 4 ln ∣ x + 2 ∣))]_{0}^{2}

Simplify

[- \frac{x ^{3}}{x + 2} - (- 3 (\frac{( x + 2 ) ^{2}}{2} - 4 (x + 2) + 4 ln ∣ x + 2 ∣))]_{0}^{2} : [- \frac{x ^{3}}{x + 2} + 3 (\frac{1}{2} (x + 2)^{2} - 4 (x + 2) + 4 ln ∣ x + 2 ∣)]_{0}^{2}

= [- \frac{x ^{3}}{x + 2} + 3 (\frac{1}{2} (x + 2)^{2} - 4 (x + 2) + 4 ln ∣ x + 2 ∣)]_{0}^{2}

Compute the boundaries:

12 ln (2) - 8

= 12 ln (2) - 8

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What is the integral from 0 to 2 of (x^3)/(x^2+4x+4) ?
The integral from 0 to 2 of (x^3)/(x^2+4x+4) is 12ln(2)-8