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

Lösung

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

2 arctan (x) + ln ∣ x ∣ + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

\int \frac{2 x}{x ^{3} + x} d x = 2 arctan (x)

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

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

Vereinfache

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

= 2 arctan (x) + ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= 2 arctan (x) + ln ∣ x ∣ + C

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

n = 1 \sum \infty \frac{1}{4 n + 2}

d er i v a t i v e g (u) = 3 u + 5 u

\int \frac{a}{x ^{a + 1}} d x

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