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

Lösung

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

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

Schritte zur Lösung

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

Faktorisiere

36 x^{2} - 1 : - (- 36 x^{2} + 1)

= \int \frac{1}{- ( - 36 x ^{2} + 1 )} d x

Entferne die Konstante:

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

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

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{11 x ^{2} + 5 x - 4}{2 x ^{3} - x ^{2}} d x

\int x^{2} x^{3} - 5 d x

\int \frac{x + 1}{x ^{4} + 3 x ^{2}} d x

\int + e^{- j wt} d t

\int \frac{1}{( 3 x - 2 ) ^{6}} d x