integral from 4 to x of 1/(1+t+t^2)

Lösung

\int_{4}^{x} \frac{1}{1 + t + t ^{2}} d t

\frac{2}{3} (arctan (\frac{2 x + 1}{3}) - arctan (33))

Schritte zur Lösung

\int_{4}^{x} \frac{1}{1 + t + t ^{2}} d t

Vervollständige das Quadrat

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

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

Wende U-Substitution an

= \int_{\frac{9}{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{9}{2}}^{x + \frac{1}{2}} \frac{1}{4 u ^{2} + 3} d u

Wende integrale Substitution an

= 4 \cdot \int_{33}^{\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_{33}^{\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)]_{33}^{\frac{2}{3} (x + \frac{1}{2})}

Vereinfache

4 \cdot \frac{1}{2 3} [arctan (v)]_{33}^{\frac{2}{3} (x + \frac{1}{2})} : \frac{2}{3} [arctan (v)]_{33}^{\frac{2 x + 1}{3}}

= \frac{2}{3} [arctan (v)]_{33}^{\frac{2 x + 1}{3}}

Berechne die Grenzen:

arctan (\frac{2 x + 1}{3}) - arctan (33)

= \frac{2}{3} (arctan (\frac{2 x + 1}{3}) - arctan (33))

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