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

Lösung

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

\frac{π}{4}

Dezimale

0.78539 \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}^{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}

\frac{\partial}{\partial φ} (ρ sin (φ) sin (θ))

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{e ^{2 x}}{x ^{2}}

x \to a lim (sin (x))

\int_{6}^{8} \frac{1}{x ^{2} - 1} d x

x \to 0 lim (x sec (x) csc (x))