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

Lösung

\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

Schritte zur Lösung

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

Wende die partielle Integration an

= - \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 ∣))

Vereinfache

- \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 ∣)

Füge eine Konstante zur Lösung hinzu

= - \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})

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

d er i v a t i v e \frac{6 e ^{s} s - 5 e ^{s}}{( 6 s + 1 ) ^{2}}