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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

\frac{2 u - 1}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} - \frac{1}{4 u ^{2} - 1}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 u}{4 u ^{2} - 1} d u - \int \frac{1}{4 u ^{2} - 1} d u

\int \frac{2 u}{4 u ^{2} - 1} d u = \frac{1}{4} ln 4 u^{2} - 1

\int \frac{1}{4 u ^{2} - 1} d u = - \frac{1}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

t an g e n t y = 2 e^{x} cos (x), (0, 2)

\int x^{π} d x

\int 2 x + 3 d x

x \to 0 lim (x^{x^{x}} - 1)

d er i v a t i v e \frac{x ^{2}}{3} + \frac{1}{x ^{2}}