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Beliebt Rechnen >

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

Lösung

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

Lösung

2 (\frac{1}{4} ln (3) - \frac{π}{12 3})

+1

Dezimale

0.24700 \dots

Schritte zur Lösung

\int_{0}^{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_{0}^{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}{12 3}

= 2 (\frac{1}{4} ln (3) - \frac{π}{12 3})

Graph

Beliebte Beispiele

(dx)/(dt)+5t^7x^9+x/t =0

\frac{d x}{d t} + 5 t^{7} x^{9} + \frac{x}{t} = 0

limit as x approaches 0-of (ln(x^2))/x

x \to 0 - lim (\frac{ln ( x ^{2} )}{x})

taylor (x^2)/(1-2x)

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2x^{''}+7x^'+3x=0

2 x^{^{''}} + 7 x^{^{'}} + 3 x = 0

tangent f(x)=e^{5x}cos(pix),\at x=0

t an g e n t f (x) = e^{5 x} cos (π x), at x = 0