integral from 0 to 1 of 1/(x^2+2x+2)

Lösung

\int_{0}^{1} \frac{1}{x ^{2} + 2 x + 2} d x

arctan (2) - \frac{π}{4}

Dezimale

0.32175 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{1}{x ^{2} + 2 x + 2} d x

Vervollständige das Quadrat

x^{2} + 2 x + 2 : (x + 1)^{2} + 1

= \int_{0}^{1} \frac{1}{( x + 1 ) ^{2} + 1} d x

Wende U-Substitution an

= \int_{1}^{2} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= [arctan (u)]_{1}^{2}

Berechne die Grenzen:

arctan (2) - \frac{π}{4}

= arctan (2) - \frac{π}{4}

\int_{0}^{1} (7 x^{2} - 6 x + 3) d x

\int_{1}^{9} t d t

\int_{0}^{π} - 4 sin (x) d x

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

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