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

Lösung

\int_{1}^{6} (\frac{3 x ^{2} + 2 x + 1}{x}) d x

ln (6) + \frac{125}{2}

Dezimale

64.29175 \dots

Schritte zur Lösung

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

Multipliziere aus

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

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

\int_{1}^{6} 3 x d x = \frac{105}{2}

\int_{1}^{6} 2 d x = 10

\int_{1}^{6} \frac{1}{x} d x = ln (6)

= \frac{105}{2} + 10 + ln (6)

Ziehe Brüche zusammen

10 + \frac{105}{2} : \frac{125}{2}

= ln (6) + \frac{125}{2}

\int_{0}^{\frac{π}{4}} 2 cos^{2} (x) d x

\int_{- 3}^{3} cos^{2} (\frac{π x}{2}) d x

\int_{x}^{y} (2 t - 3) d t

\int_{0}^{30} t e^{- s t} d t

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