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

Lösung

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

- \frac{x ^{2}}{( x - 2 ) ^{2}} + 2 ln ∣ x - 2 ∣ - \frac{8}{x - 2} - \frac{4}{( x - 2 ) ^{2}} + \frac{3}{2 ( x - 2 ) ^{2}} + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{x ^{2} + 6 x - 1}{x ^{3} - x} d x

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

\int \frac{- 8 x}{( x + 1 ) - x + 1} d x

\int \frac{2}{2 - 9 x ^{2}} d x

\int u \frac{1}{4 - 9 u ^{2}} d u