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

Lösung

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

\frac{2}{3} arctan (\frac{2 x + 1}{3}) - \frac{π}{3 3}

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Wende integrale Substitution an

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

Vereinfache

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

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

Berechne die Grenzen:

arctan (\frac{2 x + 1}{3}) - \frac{π}{6}

= \frac{2}{3} (arctan (\frac{2 x + 1}{3}) - \frac{π}{6})

Vereinfache

= \frac{2}{3} arctan (\frac{2 x + 1}{3}) - \frac{π}{3 3}

x \to 0 lim (sec (2 x))

in v erse l a pl a ce \frac{1}{s - 4}

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