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

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

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integral of (x^3)/(x^2+4x+4)

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

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

Solution steps

\int \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

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

\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 ∣))

Simplify

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

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

Add a constant to the solution

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

\int (x - 1)^{2} - 4 d x

\int (x + 1) (x + 2) d x

x \to 1 lim (\frac{1}{x + 1})

f (x) = x^{2}, x = 0, x = 2

\frac{6 e ^{s} s - 5 e ^{s}}{( 6 s + 1 ) ^{2}}

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