integral of 1/(x(4-ln^2(x)))

Lösung

\int \frac{1}{x ( 4 - ln ^{2} ( x ) )} d x

\frac{1}{4} (ln \frac{ln ( x )}{2} + 1 - ln \frac{ln ( x )}{2} - 1) + C

Schritte zur Lösung

\int \frac{1}{x ( 4 - ln ^{2} ( x ) )} d x

Wende U-Substitution an

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

Wende integrale Substitution an

= \int \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

= \frac{1}{2} \cdot \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}

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

Ersetze zurück

= \frac{1}{2} \frac{ln \frac{l n ( x )}{2} + 1}{2} - \frac{ln \frac{l n ( x )}{2} - 1}{2}

Vereinfache

\frac{1}{2} \frac{ln \frac{l n ( x )}{2} + 1}{2} - \frac{ln \frac{l n ( x )}{2} - 1}{2} : \frac{1}{4} (ln \frac{ln ( x )}{2} + 1 - ln \frac{ln ( x )}{2} - 1)

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

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (ln \frac{ln ( x )}{2} + 1 - ln \frac{ln ( x )}{2} - 1) + C

d er i v a t i v e f (x) = \frac{- 9 x}{2 x - 1}

\int \frac{cos ^{3} ( x )}{1 - sin ( x )} d x

\frac{d}{d x} (se xx)

\int_{- 1}^{- 0.5} 1.5 x^{2} d x

(1 - x^{2}) y^{^{''}} - x y^{^{'}} + y = 0