integral of 9/(x^2(x^2+25))

Lösung

\int \frac{9}{x ^{2} ( x ^{2} + 25 )} d x

9 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5})) + C

Schritte zur Lösung

\int \frac{9}{x ^{2} ( x ^{2} + 25 )} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int \frac{1}{x ^{2} ( x ^{2} + 25 )} d x

Ermittle den Partialbruch von

\frac{1}{x ^{2} ( x ^{2} + 25 )} : \frac{1}{25 x ^{2}} - \frac{1}{25 ( x ^{2} + 25 )}

= 9 \cdot \int \frac{1}{25 x ^{2}} - \frac{1}{25 ( x ^{2} + 25 )} d x

Wende die Summenregel an:

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

= 9 (\int \frac{1}{25 x ^{2}} d x - \int \frac{1}{25 ( x ^{2} + 25 )} d x)

\int \frac{1}{25 x ^{2}} d x = - \frac{1}{25 x}

\int \frac{1}{25 ( x ^{2} + 25 )} d x = \frac{1}{125} arctan (\frac{x}{5})

= 9 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5}))

Füge eine Konstante zur Lösung hinzu

= 9 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5})) + C

\frac{d x}{d t} = 6 - \frac{3 x}{500}

in v erse l a pl a ce \frac{( 6 s - 4 )}{s ^{2} - 4 s + 20}

\frac{\partial}{\partial x} (x^{\frac{2}{3}})

x \to 0 + lim (e^{x} + ln (x))

\int \frac{18}{( x + 3 ) ( x ^{2} - 9 )} d x