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integral of 1/(x(a^2-x^2))

Lösung

\int \frac{1}{x ( a ^{2} - x ^{2} )} d x

\frac{1}{a ^{2}} ln (\frac{x}{a}) - \frac{1}{2 a ^{2}} ln \frac{x ^{2}}{a ^{2}} - 1 + C

Schritte zur Lösung

\int \frac{1}{x ( a ^{2} - x ^{2} )} d x

Wende integrale Substitution an

= \int \frac{1}{u ( a ^{2} - a ^{2} u ^{2} )} d u

\frac{1}{u ( a ^{2} - a ^{2} u ^{2} )} = \frac{1}{a ^{2}} \frac{1}{u ( 1 - u ^{2} )}

= \int \frac{1}{a ^{2}} \cdot \frac{1}{u ( 1 - u ^{2} )} d u

Entferne die Konstante:

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

= \frac{1}{a ^{2}} \cdot \int \frac{1}{u ( 1 - u ^{2} )} d u

Wende U-Substitution an

= \frac{1}{a ^{2}} \cdot \int \frac{1}{2 v ( 1 - v )} d v

Entferne die Konstante:

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

= \frac{1}{a ^{2}} \cdot \frac{1}{2} \cdot \int \frac{1}{v ( 1 - v )} d v

Ermittle den Partialbruch von

\frac{1}{v ( 1 - v )} : \frac{1}{v} - \frac{1}{v - 1}

= \frac{1}{a ^{2}} \cdot \frac{1}{2} \cdot \int \frac{1}{v} - \frac{1}{v - 1} d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{a ^{2}} \cdot \frac{1}{2} (\int \frac{1}{v} d v - \int \frac{1}{v - 1} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= \frac{1}{a ^{2}} \cdot \frac{1}{2} (ln ∣ v ∣ - ln ∣ v - 1 ∣)

Ersetze zurück

= \frac{1}{a ^{2}} \cdot \frac{1}{2} (ln (\frac{x}{a})^{2} - ln (\frac{x}{a})^{2} - 1)

Vereinfache

\frac{1}{a ^{2}} \cdot \frac{1}{2} (ln (\frac{x}{a})^{2} - ln (\frac{x}{a})^{2} - 1) : \frac{1}{a ^{2}} ln (\frac{x}{a}) - \frac{1}{2 a ^{2}} ln \frac{x ^{2}}{a ^{2}} - 1

= \frac{1}{a ^{2}} ln (\frac{x}{a}) - \frac{1}{2 a ^{2}} ln \frac{x ^{2}}{a ^{2}} - 1

Füge eine Konstante zur Lösung hinzu

= \frac{1}{a ^{2}} ln (\frac{x}{a}) - \frac{1}{2 a ^{2}} ln \frac{x ^{2}}{a ^{2}} - 1 + C

\int \frac{1}{4 + 9 x ^{2}} d x

s l o p e y = 8 x - x^{2} (1.7)

x \to 0 lim (e^{x} + \frac{1}{x})

\frac{d}{d x} (0.8 x^{1.3})

x \to 3 - lim (- x + 5)