integral of xln((x+2)/(x+1))

Lösung

\int x ln (\frac{x + 2}{x + 1}) d x

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

Schritte zur Lösung

\int x ln (\frac{x + 2}{x + 1}) d x

Wende die partielle Integration an

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

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{4 x}{y} d x

\frac{d}{d x} (x \cdot e^{- x})

t an g e n t f (x) = 2 x^{2} - 1, at x = - 1

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

\int 3 x^{4} d x