integral from 2 to 3 of 1/(x^2-6x+10)

Lösung

\int_{2}^{3} \frac{1}{x ^{2} - 6 x + 10} d x

\frac{π}{4}

Dezimale

0.78539 \dots

Schritte zur Lösung

\int_{2}^{3} \frac{1}{x ^{2} - 6 x + 10} d x

Vervollständige das Quadrat

x^{2} - 6 x + 10 : (x - 3)^{2} + 1

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

Wende U-Substitution an

= \int_{- 1}^{0} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

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

= [arctan (u)]_{- 1}^{0}

Berechne die Grenzen:

\frac{π}{4}

= \frac{π}{4}

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

\int_{- 3}^{0} (\frac{y}{3} + y + 4) d y

\int_{0}^{9 π} 3 + sin (x) d x

\int_{- π}^{π} sin (x) \cdot sin (n x) d x

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