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

Lösung

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

- \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣) + C

Schritte zur Lösung

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

Faktorisiere

4 x^{2} - 1 : - (- 4 x^{2} + 1)

= \int \frac{1}{- ( - 4 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}{- 4 x ^{2} + 1} d x

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Setze in

u = 2 x

ein

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

Vereinfache

- \frac{1}{2} (\frac{ln ∣ 2 x + 1 ∣}{2} - \frac{ln ∣ 2 x - 1 ∣}{2}) : - \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣)

= - \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣) + C

\frac{\partial}{\partial x} (sin (6 x y + 7 yz + 9 x z))

\int (4 x - \frac{2}{x} + 5 sin (x)) d x

x \to \infty lim ((4)^{x})

f^{^{'}} (x) = sec (2 x) tan (2 x) - \frac{4}{x ^{2}}

n = 0 \sum \infty x^{n} \frac{1}{n}