integral of (2x-3)/(4x^2-1)

Lösung

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

\frac{1}{4} ln 4 x^{2} - 1 + \frac{3}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int \frac{3}{4 x ^{2} - 1} d x = - \frac{3}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣)

= \frac{1}{4} ln 4 x^{2} - 1 - (- \frac{3}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣))

Vereinfache

= \frac{1}{4} ln 4 x^{2} - 1 + \frac{3}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} ln 4 x^{2} - 1 + \frac{3}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣) + C

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