integral of 1/(4-9x^2)

Lösung

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

\frac{1}{12} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1) + C

Schritte zur Lösung

\int \frac{1}{4 - 9 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{3}{2} x

ein

= \frac{1}{6} (\frac{ln \frac{3}{2} x + 1}{2} - \frac{ln \frac{3}{2} x - 1}{2})

Vereinfache

\frac{1}{6} (\frac{ln \frac{3}{2} x + 1}{2} - \frac{ln \frac{3}{2} x - 1}{2}) : \frac{1}{12} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1)

= \frac{1}{12} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{12} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1) + C

f (x) = 2 x + 3

\frac{d}{d x} (\frac{2}{3 - x})

\frac{d}{d x} (sec (tan^{2} (x^{4})))

d y = (x y^{2} + 3 x y) d x

y^{^{''}} + 4 y^{^{'}} + 4 y = 25 cos (x)