integral from 1 to 4 of (2x+1)/(x^2+x-1)

Lösung

\int_{1}^{4} \frac{2 x + 1}{x ^{2} + x - 1} d x

ln (19)

Dezimale

2.94443 \dots

Schritte zur Lösung

\int_{1}^{4} \frac{2 x + 1}{x ^{2} + x - 1} d x

Multipliziere aus

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

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}^{4} \frac{2 x}{x ^{2} + x - 1} d x + \int_{1}^{4} \frac{1}{x ^{2} + x - 1} d x

\int_{1}^{4} \frac{2 x}{x ^{2} + x - 1} d x = ln (19) + \frac{1}{5} ln (\frac{83 - 33 5}{38})

\int_{1}^{4} \frac{1}{x ^{2} + x - 1} d x = - \frac{ln ( \frac{83 - 33 5}{38} )}{5}

= ln (19) + \frac{1}{5} ln (\frac{83 - 33 5}{38}) - \frac{ln ( \frac{83 - 33 5}{38} )}{5}

Vereinfache

ln (19) + \frac{1}{5} ln (\frac{83 - 33 5}{38}) - \frac{ln ( \frac{83 - 33 5}{38} )}{5} : ln (19)

= ln (19)