integral of 1/(36-x^2)

Lösung

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

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

Schritte zur Lösung

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

Wende integrale Substitution an

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

Nutze das gemeinsame Integral :

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

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

Setze in

u = \frac{x}{6}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int sin^{3} (6 x) cos^{- 2} (6 x) d x

\int \frac{I n x}{x} d x

y^{^{''}} + 9 y^{^{'}} + 8 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int_{1}^{\infty} x e^{- 9 x} d x

\int sin^{3} (x) \cdot cos^{3} (x) d x