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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

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

\frac{d}{d x} (7 x^{5})

d er i v a t i v e \frac{x - 3 x x}{x}

\frac{d}{d x} (\frac{cos ( x )}{- 2 sin ( 2 x )})

\frac{\partial}{\partial r} (z)