integral from 1 to 2 of 1/(x(x-5))

Lösung

\int_{1}^{2} \frac{1}{x ( x - 5 )} d x

\frac{- 3 ln ( 2 ) + ln ( 3 )}{5}

Dezimale

- 0.19616 \dots

Schritte zur Lösung

\int_{1}^{2} \frac{1}{x ( x - 5 )} d x

Ermittle den Partialbruch von

\frac{1}{x ( x - 5 )} : - \frac{1}{5 x} + \frac{1}{5 ( x - 5 )}

= \int_{1}^{2} - \frac{1}{5 x} + \frac{1}{5 ( x - 5 )} 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

= - \int_{1}^{2} \frac{1}{5 x} d x + \int_{1}^{2} \frac{1}{5 ( x - 5 )} d x

\int_{1}^{2} \frac{1}{5 x} d x = \frac{1}{5} ln (2)

\int_{1}^{2} \frac{1}{5 ( x - 5 )} d x = \frac{1}{5} (ln (3) - 2 ln (2))

= - \frac{1}{5} ln (2) + \frac{1}{5} (ln (3) - 2 ln (2))

Vereinfache

- \frac{1}{5} ln (2) + \frac{1}{5} (ln (3) - 2 ln (2)) : \frac{- 3 ln ( 2 ) + ln ( 3 )}{5}

= \frac{- 3 ln ( 2 ) + ln ( 3 )}{5}

\int \frac{1}{x 2 + 9 x ^{2}} d x

\frac{d}{d x} (4 x^{\frac{3}{4}})

\int θ arcsec (9 θ) d θ

d er i v a t i v e F (r) = \frac{4}{r ^{3}}

x^{3} y^{^{'''}} + x y^{^{'}} - y = x^{2}