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

Lösung

\int_{1}^{2} \frac{1 + y}{1 + y ^{2}} d y

arctan (2) - \frac{π}{4} + \frac{1}{2} (ln (5) - ln (2))

Dezimale

0.77989 \dots

Schritte zur Lösung

\int_{1}^{2} \frac{1 + y}{1 + y ^{2}} d y

Multipliziere aus

\frac{1 + y}{1 + y ^{2}} : \frac{1}{1 + y ^{2}} + \frac{y}{1 + y ^{2}}

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}{1 + y ^{2}} d y + \int_{1}^{2} \frac{y}{1 + y ^{2}} d y

\int_{1}^{2} \frac{1}{1 + y ^{2}} d y = arctan (2) - \frac{π}{4}

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

= arctan (2) - \frac{π}{4} + \frac{1}{2} (ln (5) - ln (2))

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