integral of (2x^3+4x^2+3x-1)/(x^2+2x+1)

Lösung

\int \frac{2 x ^{3} + 4 x ^{2} + 3 x - 1}{x ^{2} + 2 x + 1} d x

x^{2} + 4 x + \frac{6}{x + 1} + ln ∣ x + 1 ∣ - \frac{4 x ^{2}}{x + 1} + 3 + C

Schritte zur Lösung

\int \frac{2 x ^{3} + 4 x ^{2} + 3 x - 1}{x ^{2} + 2 x + 1} d x

Multipliziere aus

\frac{2 x ^{3} + 4 x ^{2} + 3 x - 1}{x ^{2} + 2 x + 1} : \frac{2 x ^{3}}{x ^{2} + 2 x + 1} + \frac{4 x ^{2}}{x ^{2} + 2 x + 1} + \frac{3 x}{x ^{2} + 2 x + 1} - \frac{1}{x ^{2} + 2 x + 1}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 x ^{3}}{x ^{2} + 2 x + 1} d x + \int \frac{4 x ^{2}}{x ^{2} + 2 x + 1} d x + \int \frac{3 x}{x ^{2} + 2 x + 1} d x - \int \frac{1}{x ^{2} + 2 x + 1} d x

\int \frac{2 x ^{3}}{x ^{2} + 2 x + 1} d x = x^{2} - 4 x + \frac{2}{x + 1} + 6 ln ∣ x + 1 ∣ - 5

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

\int \frac{3 x}{x ^{2} + 2 x + 1} d x = 3 (ln ∣ x + 1 ∣ + \frac{1}{x + 1})

\int \frac{1}{x ^{2} + 2 x + 1} d x = - \frac{1}{x + 1}

= x^{2} - 4 x + \frac{2}{x + 1} + 6 ln ∣ x + 1 ∣ - 5 + 4 (- \frac{x ^{2}}{x + 1} + 2 (x + 1 - ln ∣ x + 1 ∣)) + 3 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - (- \frac{1}{x + 1})

Vereinfache

x^{2} - 4 x + \frac{2}{x + 1} + 6 ln ∣ x + 1 ∣ - 5 + 4 (- \frac{x ^{2}}{x + 1} + 2 (x + 1 - ln ∣ x + 1 ∣)) + 3 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - (- \frac{1}{x + 1}) : x^{2} + 4 x + \frac{6}{x + 1} + ln ∣ x + 1 ∣ - \frac{4 x ^{2}}{x + 1} + 3

= x^{2} + 4 x + \frac{6}{x + 1} + ln ∣ x + 1 ∣ - \frac{4 x ^{2}}{x + 1} + 3

Füge eine Konstante zur Lösung hinzu

= x^{2} + 4 x + \frac{6}{x + 1} + ln ∣ x + 1 ∣ - \frac{4 x ^{2}}{x + 1} + 3 + C

t an g e n t f (x) = (x - 4) (x^{2} + 8), (1, - 27)

x \to 1 - lim (- x^{2} + 5 x + (- 4))

n = 3 \sum \infty \frac{2}{n ln ^{2} ( n )}

a re a y = x, y = 3 x

\int \frac{x}{( 5 x ^{2} + 2 ) ^{6}} d x