integral ((x^2+x+1))/((x^2-x))

Lösung

integral

\frac{( x ^{2} + x + 1 )}{( x ^{2} - x )}

x + 3 ln ∣ x - 1 ∣ - ln ∣ x ∣ - 1 + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

= x - 1 + ln ∣ x - 1 ∣ + ln ∣ x - 1 ∣ - ln ∣ x ∣ + ln ∣ x - 1 ∣

Vereinfache

= x + 3 ln ∣ x - 1 ∣ - ln ∣ x ∣ - 1

Füge eine Konstante zur Lösung hinzu

= x + 3 ln ∣ x - 1 ∣ - ln ∣ x ∣ - 1 + C

in t e g r a l \frac{( x ^{2} + 2 x - 3 )}{x ^{4}}

in t e g r a l y (1 - y^{2})^{\frac{1}{2}}

in t e g r a l 2 e^{- x}

in t e g r a l \frac{1}{( 1 + x ^{2} ) ^{0.5}}

in t e g r a l \frac{- 1}{y + 1}