integral from 1 to 2 of 1/(x+x^3)

Lösung

\int_{1}^{2} \frac{1}{x + x ^{3}} d x

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

Dezimale

0.23500 \dots

Schritte zur Lösung

\int_{1}^{2} \frac{1}{x + x ^{3}} d x

Ermittle den Partialbruch von

\frac{1}{x + x ^{3}} : \frac{1}{x} - \frac{x}{x ^{2} + 1}

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

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

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

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

Vereinfache

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

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

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

\int_{0}^{2} 2 x (x^{2} + 4)^{2} d x

\int_{1}^{4} (- x + \frac{4}{x}) d x

\int_{1}^{10} \frac{5}{x} d x

\int_{0}^{2} (x^{2} - 10 x + 16) d x