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integral from 0 to 1 of (56)/(8x^2+9x+1)

Lösung

\int_{0}^{1} \frac{56}{8 x ^{2} + 9 x + 1} d x

- 8 ln (\frac{2}{9})

Dezimale

12.03261 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{56}{8 x ^{2} + 9 x + 1} d x

Entferne die Konstante:

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

= 56 \cdot \int_{0}^{1} \frac{1}{8 x ^{2} + 9 x + 1} d x

Vervollständige das Quadrat

8 x^{2} + 9 x + 1 : 8 (x + \frac{9}{16})^{2} - \frac{49}{32}

= 56 \cdot \int_{0}^{1} \frac{1}{8 ( x + \frac{9}{16} ) ^{2} - \frac{49}{32}} d x

Wende U-Substitution an

= 56 \cdot \int_{\frac{9}{16}}^{\frac{25}{16}} \frac{32}{256 u ^{2} - 49} d u

Entferne die Konstante:

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

= 56 \cdot 32 \cdot \int_{\frac{9}{16}}^{\frac{25}{16}} \frac{1}{256 u ^{2} - 49} d u

Wende integrale Substitution an

= 56 \cdot 32 \cdot \int_{\frac{9}{7}}^{\frac{25}{7}} \frac{1}{112 ( v ^{2} - 1 )} d v

Entferne die Konstante:

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

= 56 \cdot 32 \cdot \frac{1}{112} \cdot \int_{\frac{9}{7}}^{\frac{25}{7}} \frac{1}{v ^{2} - 1} d v

Faktorisiere

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

= 56 \cdot 32 \cdot \frac{1}{112} \cdot \int_{\frac{9}{7}}^{\frac{25}{7}} \frac{1}{- ( - v ^{2} + 1 )} d v

Entferne die Konstante:

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

= 56 \cdot 32 \cdot \frac{1}{112} (- \int_{\frac{9}{7}}^{\frac{25}{7}} \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}

= 56 \cdot 32 \cdot \frac{1}{112} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{9}{7}}^{\frac{25}{7}})

Vereinfache

56 \cdot 32 \cdot \frac{1}{112} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{9}{7}}^{\frac{25}{7}}) : - 16 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{9}{7}}^{\frac{25}{7}}

= - 16 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{9}{7}}^{\frac{25}{7}}

Berechne die Grenzen:

\frac{ln ( \frac{32}{7} ) - ln ( \frac{18}{7} ) - ln ( \frac{16}{7} ) + ln ( \frac{2}{7} )}{2}

= - 16 \cdot \frac{ln ( \frac{32}{7} ) - ln ( \frac{18}{7} ) - ln ( \frac{16}{7} ) + ln ( \frac{2}{7} )}{2}

Vereinfache

= - 8 ln (\frac{2}{9})

\int \frac{x + 1}{4 x} (x - 1) d x

\int_{1}^{3} 13 r^{3} ln (r) d r

n = 0 \sum \infty 5 n

\int \frac{( 7 ( 1 - tan ^{2} ( x ) ) )}{sec ^{2} ( x )} d x

\frac{\partial}{\partial x} (ln (4 x^{2} + 7 y^{2} + 4))