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integral from 3 to 4 of 1/(x^2-3x+2)

Lösung

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

- ln (\frac{3}{4})

Dezimale

0.28768 \dots

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

= \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{4}{4 u ^{2} - 1} d u

Entferne die Konstante:

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

= 4 \cdot \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{4 u ^{2} - 1} d u

Faktorisiere

4 u^{2} - 1 : - (- 4 u^{2} + 1)

= 4 \cdot \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{- ( - 4 u ^{2} + 1 )} d u

Entferne die Konstante:

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

= 4 (- \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{- 4 u ^{2} + 1} d u)

Wende integrale Substitution an

= 4 (- \int_{3}^{5} \frac{1}{2 ( - v ^{2} + 1 )} d v)

Entferne die Konstante:

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

= 4 (- \frac{1}{2} \cdot \int_{3}^{5} \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 (- \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{3}^{5})

Vereinfache

4 (- \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{3}^{5}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{3}^{5}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{3}^{5}

Berechne die Grenzen:

\frac{ln ( 6 ) - 3 ln ( 2 )}{2}

= - 2 \cdot \frac{ln ( 6 ) - 3 ln ( 2 )}{2}

Vereinfache

= - ln (\frac{3}{4})

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 4)^{2} d x

\int_{0}^{2} x^{3} + x d x

\int_{2}^{6} \frac{1}{x ^{2}} d x

\int_{0}^{6} e^{\frac{x}{2}} d x

\int_{- \frac{π}{2}}^{π} 7 cos (x) d x