integral of 1/(6-36x^2)

Lösung

\int \frac{1}{6 - 36 x ^{2}} d x

\frac{1}{12 6} (ln 6 x + 1 - ln 6 x - 1) + C

Schritte zur Lösung

\int \frac{1}{6 - 36 x ^{2}} d x

Wende integrale Substitution an

= \int \frac{1}{6 ( 6 - 6 u ^{2} )} d u

Entferne die Konstante:

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

= \frac{1}{6} \cdot \int \frac{1}{6 - 6 u ^{2}} d u

\frac{1}{6 - 6 u ^{2}} = \frac{1}{6} \frac{1}{1 - u ^{2}}

= \frac{1}{6} \cdot \int \frac{1}{6} \cdot \frac{1}{1 - u ^{2}} d u

Entferne die Konstante:

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

= \frac{1}{6} \cdot \frac{1}{6} \cdot \int \frac{1}{1 - u ^{2}} d u

Nutze das gemeinsame Integral :

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{6} \cdot \frac{1}{6} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Setze in

u = \frac{6}{6} x

ein

= \frac{1}{6} \cdot \frac{1}{6} \frac{ln \frac{6}{6} x + 1}{2} - \frac{ln \frac{6}{6} x - 1}{2}

Vereinfache

\frac{1}{6} \cdot \frac{1}{6} \frac{ln \frac{6}{6} x + 1}{2} - \frac{ln \frac{6}{6} x - 1}{2} : \frac{1}{12 6} (ln 6 x + 1 - ln 6 x - 1)

= \frac{1}{12 6} (ln 6 x + 1 - ln 6 x - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{12 6} (ln 6 x + 1 - ln 6 x - 1) + C

d er i v a t i v e \frac{2 - sec ( x )}{tan ( x )}

\int e^{x} 15 + e^{x} d x

\frac{d}{d x} (\frac{3}{8} (x^{\frac{4}{3}} - 2 x^{\frac{2}{3}}))

\frac{d}{d x} ((x^{3} + 2)^{2} (x^{4} + 4)^{4})

y^{^{'}} (t) = 0.3 y \cdot (1 - \frac{y}{2500}) - \frac{y}{4}, y (0) = 600