integral of 1/(25-4x^2)

Lösung

\int \frac{1}{25 - 4 x ^{2}} d x

\frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) + C

Schritte zur Lösung

\int \frac{1}{25 - 4 x ^{2}} d x

Wende integrale Substitution an

= \int \frac{1}{10 ( - u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{10} \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}{10} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Setze in

u = \frac{2}{5} x

ein

= \frac{1}{10} (\frac{ln \frac{2}{5} x + 1}{2} - \frac{ln \frac{2}{5} x - 1}{2})

Vereinfache

\frac{1}{10} (\frac{ln \frac{2}{5} x + 1}{2} - \frac{ln \frac{2}{5} x - 1}{2}) : \frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1)

= \frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) + C

d er i v a t i v e \frac{6 x}{x + 5}

x \to 8 lim (\frac{( x - 8 )}{x - 8})

\int \frac{4}{u} d u

x \to \infty lim (k^{x} c)

\int_{- 1}^{3} (3 x + 1) d x