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

Lösung

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

2 x^{2} - 1 - ln x^{2} - 1 + x + C

Schritte zur Lösung

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

Multipliziere aus

\frac{2 x - 1}{x ^{2} - 1} : \frac{2 x}{x ^{2} - 1} - \frac{1}{x ^{2} - 1}

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

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

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

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

= 2 x^{2} - 1 - ln x^{2} - 1 + x

Füge eine Konstante zur Lösung hinzu

= 2 x^{2} - 1 - ln x^{2} - 1 + x + C

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

(2 sec^{2} (x) tan (x))^{^{'}}

\int \frac{1}{( x + 2 ) x ^{2} + 4 x - 12} d x

ma c l a u r in \frac{1}{1 + x ^{2}}

x \to 0 - lim (6 x - 1)