integral of ((11x-7))/(2x^2-3x-2)

Lösung

\int \frac{( 11 x - 7 )}{2 x ^{2} - 3 x - 2} d x

\frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣ + C

Schritte zur Lösung

\int \frac{11 x - 7}{2 x ^{2} - 3 x - 2} d x

Multipliziere aus

\frac{11 x - 7}{2 x ^{2} - 3 x - 2} : \frac{11 x}{2 x ^{2} - 3 x - 2} - \frac{7}{2 x ^{2} - 3 x - 2}

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{11 x}{2 x ^{2} - 3 x - 2} d x - \int \frac{7}{2 x ^{2} - 3 x - 2} d x

\int \frac{11 x}{2 x ^{2} - 3 x - 2} d x = 11 (\frac{1}{10} ln ∣ 2 x + 1 ∣ + \frac{2}{5} ln ∣ x - 2 ∣)

\int \frac{7}{2 x ^{2} - 3 x - 2} d x = 7 (- \frac{1}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣)

= 11 (\frac{1}{10} ln ∣ 2 x + 1 ∣ + \frac{2}{5} ln ∣ x - 2 ∣) - 7 (- \frac{1}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣)

Vereinfache

11 (\frac{1}{10} ln ∣ 2 x + 1 ∣ + \frac{2}{5} ln ∣ x - 2 ∣) - 7 (- \frac{1}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣) : \frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣

= \frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣ + C

\int \frac{16 x}{8 x ^{2} + 2} d x

\int \frac{2 sin ^{2} ( x )}{6} d x

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

\int \frac{cos ( \frac{π}{x ^{4}} )}{x ^{5}} d x

\int \frac{1}{( arctan ( y ) ) ( 1 + y ^{2} )} d y