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integral of 1/(x^2-3x-15)

Lösung

\int \frac{1}{x ^{2} - 3 x - 15} d x

- \frac{1}{69} (ln \frac{2}{69} x - \frac{3}{23} + 1 - ln \frac{2}{69} x - \frac{3}{23} - 1) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} - 3 x - 15} d x

Vervollständige das Quadrat

x^{2} - 3 x - 15 : (x - \frac{3}{2})^{2} - \frac{69}{4}

= \int \frac{1}{( x - \frac{3}{2} ) ^{2} - \frac{69}{4}} d x

Wende U-Substitution an

= \int \frac{4}{4 u ^{2} - 69} d u

Entferne die Konstante:

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

= 4 \cdot \int \frac{1}{4 u ^{2} - 69} d u

Wende integrale Substitution an

= 4 \cdot \int \frac{1}{2 69 ( 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 69} \cdot \int \frac{1}{v ^{2} - 1} d v

Faktorisiere

v^{2} - 1 : - (- v^{2} + 1)

= 4 \cdot \frac{1}{2 69} \cdot \int \frac{1}{- ( - 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 69} (- \int \frac{1}{- v ^{2} + 1} d v)

Nutze das gemeinsame Integral :

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 4 \cdot \frac{1}{2 69} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Ersetze zurück

= 4 \cdot \frac{1}{2 69} - \frac{ln \frac{2}{69} ( x - \frac{3}{2} ) + 1}{2} - \frac{ln \frac{2}{69} ( x - \frac{3}{2} ) - 1}{2}

Vereinfache

4 \cdot \frac{1}{2 69} - \frac{ln \frac{2}{69} ( x - \frac{3}{2} ) + 1}{2} - \frac{ln \frac{2}{69} ( x - \frac{3}{2} ) - 1}{2} : - \frac{1}{69} (ln \frac{2}{69} x - \frac{3}{23} + 1 - ln \frac{2}{69} x - \frac{3}{23} - 1)

= - \frac{1}{69} (ln \frac{2}{69} x - \frac{3}{23} + 1 - ln \frac{2}{69} x - \frac{3}{23} - 1)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{69} (ln \frac{2}{69} x - \frac{3}{23} + 1 - ln \frac{2}{69} x - \frac{3}{23} - 1) + C

\int 40 + 6 x - x^{2} d x

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\frac{d}{d x} (tan (x) d x)

\int e^{\frac{2 x}{5}} d x

t a y l or \frac{1}{1 - x}, 20