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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

\frac{x + 1}{x ( x ^{2} + 2 )} : \frac{x}{x ( x ^{2} + 2 )} + \frac{1}{x ( x ^{2} + 2 )}

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{x}{x ^{2} - 14 x + 45} d x

\int a^{2} + t^{2} d t

\int (tan (x) - sec (x))^{2} d x

\int tan^{5} (\frac{x}{18}) d x

\int \frac{y}{x + 1} d x