integral from 1/4 to 1 of (y+6)/(y^2+3y)

Lösung

\int_{\frac{1}{4}}^{1} \frac{y + 6}{y ^{2} + 3 y} d y

2 ln (2) + ln (\frac{13}{4})

Dezimale

2.56494 \dots

Schritte zur Lösung

\int_{\frac{1}{4}}^{1} \frac{y + 6}{y ^{2} + 3 y} d y

Ermittle den Partialbruch von

\frac{y + 6}{y ^{2} + 3 y} : \frac{2}{y} - \frac{1}{y + 3}

= \int_{\frac{1}{4}}^{1} \frac{2}{y} - \frac{1}{y + 3} d y

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{1}{4}}^{1} \frac{2}{y} d y - \int_{\frac{1}{4}}^{1} \frac{1}{y + 3} d y

\int_{\frac{1}{4}}^{1} \frac{2}{y} d y = 4 ln (2)

\int_{\frac{1}{4}}^{1} \frac{1}{y + 3} d y = 2 ln (2) - ln (\frac{13}{4})

= 4 ln (2) - (2 ln (2) - ln (\frac{13}{4}))

Vereinfache

= 2 ln (2) + ln (\frac{13}{4})

\int_{1}^{2} \frac{6 x}{3 x ^{2} - 2} d x

\int_{1}^{\infty} \frac{9}{x ^{3}} d x

\int_{4}^{5} \frac{2}{( x - 3 ) ^{3}} d x

\int_{4}^{5} \frac{1}{3 x ^{2} - 3} d x

\int_{0}^{t} x e^{- 4 x} d x