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

Lösung

\int \frac{( x - 4 )}{x ^{2} + 3 x + 2} d x

- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ + 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int \frac{4}{x ^{2} + 3 x + 2} d x = - 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣)

= - ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ - (- 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣))

Vereinfache

= - ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ + 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣)

Füge eine Konstante zur Lösung hinzu

= - ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ + 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣) + C

\int (sin (3 x) + cos (5 x) - sin (\frac{x}{2})) d x

\int \frac{1}{y - 1} d y

\int tan^{2} (x) + cot^{2} (x) d x

\int \frac{3 x ^{2}}{cos ^{2} ( x ^{3} )} d x

\int \frac{8}{5 x 25 x ^{2} - 25} d x