integral from 1 to x of 1/(6+t+3t^2)

Lösung

\int_{1}^{x} \frac{1}{6 + t + 3 t ^{2}} d t

\frac{2 ( arctan ( \frac{6 x + 1}{71} ) - arctan ( \frac{7}{71} ) )}{71}

Schritte zur Lösung

\int_{1}^{x} \frac{1}{6 + t + 3 t ^{2}} d t

Vervollständige das Quadrat

6 + t + 3 t^{2} : 3 (t + \frac{1}{6})^{2} + \frac{71}{12}

= \int_{1}^{x} \frac{1}{3 ( t + \frac{1}{6} ) ^{2} + \frac{71}{12}} d t

Wende U-Substitution an

= \int_{\frac{7}{6}}^{x + \frac{1}{6}} \frac{12}{36 u ^{2} + 71} d u

Entferne die Konstante:

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

= 12 \cdot \int_{\frac{7}{6}}^{x + \frac{1}{6}} \frac{1}{36 u ^{2} + 71} d u

Wende integrale Substitution an

= 12 \cdot \int_{\frac{7}{71}}^{\frac{6}{71} (x + \frac{1}{6})} \frac{1}{6 71 ( v ^{2} + 1 )} d v

Entferne die Konstante:

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

= 12 \cdot \frac{1}{6 71} \cdot \int_{\frac{7}{71}}^{\frac{6}{71} (x + \frac{1}{6})} \frac{1}{v ^{2} + 1} d v

Nutze das gemeinsame Integral :

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

= 12 \cdot \frac{1}{6 71} [arctan (v)]_{\frac{7}{71}}^{\frac{6}{71} (x + \frac{1}{6})}

Vereinfache

12 \cdot \frac{1}{6 71} [arctan (v)]_{\frac{7}{71}}^{\frac{6}{71} (x + \frac{1}{6})} : \frac{2}{71} [arctan (v)]_{\frac{7}{71}}^{\frac{6 x + 1}{71}}

= \frac{2}{71} [arctan (v)]_{\frac{7}{71}}^{\frac{6 x + 1}{71}}

Berechne die Grenzen:

arctan (\frac{6 x + 1}{71}) - arctan (\frac{7}{71})

= \frac{2}{71} (arctan (\frac{6 x + 1}{71}) - arctan (\frac{7}{71}))

Vereinfache

= \frac{2 ( arctan ( \frac{6 x + 1}{71} ) - arctan ( \frac{7}{71} ) )}{71}

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