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integral from 0 to 1 of 1/(x(2x+5))

Lösung

\int_{0}^{1} \frac{1}{x ( 2 x + 5 )} d x

divergiert

Schritte zur Lösung

\int_{0}^{1} \frac{1}{x ( 2 x + 5 )} d x

Vervollständige das Quadrat

x (2 x + 5) : 2 (x + \frac{5}{4})^{2} - \frac{25}{8}

= \int_{0}^{1} \frac{1}{2 ( x + \frac{5}{4} ) ^{2} - \frac{25}{8}} d x

Wende U-Substitution an

= \int_{\frac{5}{4}}^{\frac{9}{4}} \frac{8}{16 u ^{2} - 25} d u

Entferne die Konstante:

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

= 8 \cdot \int_{\frac{5}{4}}^{\frac{9}{4}} \frac{1}{16 u ^{2} - 25} d u

Wende integrale Substitution an

= 8 \cdot \int_{1}^{\frac{9}{5}} \frac{1}{20 ( v ^{2} - 1 )} d v

Entferne die Konstante:

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

= 8 \cdot \frac{1}{20} \cdot \int_{1}^{\frac{9}{5}} \frac{1}{v ^{2} - 1} d v

Faktorisiere

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

= 8 \cdot \frac{1}{20} \cdot \int_{1}^{\frac{9}{5}} \frac{1}{- ( - v ^{2} + 1 )} d v

Entferne die Konstante:

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

= 8 \cdot \frac{1}{20} (- \int_{1}^{\frac{9}{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}

= 8 \cdot \frac{1}{20} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{1}^{\frac{9}{5}})

Vereinfache

8 \cdot \frac{1}{20} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{1}^{\frac{9}{5}}) : - \frac{2}{5} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{1}^{\frac{9}{5}}

= - \frac{2}{5} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{1}^{\frac{9}{5}}

Berechne die Grenzen:

divergiert

= divergiert

\frac{\partial}{\partial x} (tan^{2} (x + y))

\int - 9.8 t d t

n = 0 \sum \infty (\frac{5}{2})^{n}

n = 1 \sum \infty \frac{1}{3 ^{n} + 5}

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