integral of (x+1)/(2x^2+4x+5)

Lösung

\int \frac{x + 1}{2 x ^{2} + 4 x + 5} d x

\frac{1}{4} ln 2 x^{2} + 4 x + 5 + C

Schritte zur Lösung

\int \frac{x + 1}{2 x ^{2} + 4 x + 5} d x

Multipliziere aus

\frac{x + 1}{2 x ^{2} + 4 x + 5} : \frac{x}{2 x ^{2} + 4 x + 5} + \frac{1}{2 x ^{2} + 4 x + 5}

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 ^{2} + 4 x + 5} d x + \int \frac{1}{2 x ^{2} + 4 x + 5} d x

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

\int \frac{1}{2 x ^{2} + 4 x + 5} d x = \frac{1}{6} arctan (\frac{2}{3} (x + 1))

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

Vereinfache

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

= \frac{1}{4} ln 2 x^{2} + 4 x + 5

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} ln 2 x^{2} + 4 x + 5 + C

\int \frac{16}{x ^{2}} d x

d er i v a t i v e 2 e^{x} + \frac{4}{3 x}

\int_{0}^{4} x^{2} 16 - x^{2} d x

\frac{d y}{d x} - x y^{2} = 2 x y

n = 0 \sum \infty \frac{1}{2 ^{2 n}}