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

Lösung

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

\frac{2 π}{3 3}

Dezimale

1.20919 \dots

Schritte zur Lösung

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

Vervollständige das Quadrat

x^{2} - x + 1 : (x - \frac{1}{2})^{2} + \frac{3}{4}

= \int_{0}^{1} \frac{1}{( x - \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

Wende U-Substitution an

= \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{4}{4 u ^{2} + 3} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{4 u ^{2} + 3} d u

Wende integrale Substitution an

= 4 \cdot \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{1}{2 3 ( v ^{2} + 1 )} d v

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{2 3} \cdot \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{1}{v ^{2} + 1} d v

Nutze das gemeinsame Integral :

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

= 4 \cdot \frac{1}{2 3} [arctan (v)]_{- \frac{1}{3}}^{\frac{1}{3}}

Vereinfache

4 \cdot \frac{1}{2 3} [arctan (v)]_{- \frac{1}{3}}^{\frac{1}{3}} : \frac{2}{3} [arctan (v)]_{- \frac{1}{3}}^{\frac{1}{3}}

= \frac{2}{3} [arctan (v)]_{- \frac{1}{3}}^{\frac{1}{3}}

Berechne die Grenzen:

\frac{π}{3}

= \frac{2}{3} \cdot \frac{π}{3}

Vereinfache

= \frac{2 π}{3 3}

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

\int_{1}^{5} x^{2} - 4 x - 12 d x

\int_{0}^{2} \frac{1}{t - 2} d t

\int_{2}^{8} x^{2} + x + 1 + \frac{1}{x} d x

\int_{0}^{5} (- 3 x - 5 x^{2}) d x