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

Lösung

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

ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int \frac{2 x}{x ^{2} + 4} d x = ln x^{2} + 4

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

= ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})

Füge eine Konstante zur Lösung hinzu

= ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

y^{^{''}} + 12 y^{^{'}} + 24 y = 4381 e^{2 t} cos (13 t)

\int e^{x} sec (e^{x}) d x

\int tan^{5} (7 x) d x

\frac{d}{d x} (\frac{10}{x - 6})

f (x) = \frac{x}{x ^{2} - x + 1}