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

Lösung

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

2 (\frac{1}{4} ln (3) - \frac{π}{4 3})

Dezimale

- 0.35759 \dots

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

\frac{2 u - 1}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} - \frac{1}{4 u ^{2} + 3}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= 2 (\frac{1}{4} ln (3) - \frac{π}{4 3})

\int_{2}^{5} \frac{1}{( x - 1 ) ( x + 2 )} d x

\int_{0}^{π} (2 sin (x)) d x

\int_{0}^{3} x e^{x^{2}} d x

\int_{0}^{1} \frac{1}{x ^{4} - 1} d x

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