integral of (4x^3+x+1)/((x-1)(x^2+x+1))

Lösung

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

4 x + 2 ln ∣ x - 1 ∣ - ln 4 x^{2} + 4 x + 4 + \frac{2}{3} arctan (\frac{1}{3} (2 x + 1)) - 23 arctan (\frac{1}{3} (2 x + 1)) + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int 4 d x = 4 x

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

\int \frac{- 2 x - 3}{x ^{2} + x + 1} d x = - ln 4 x^{2} + 4 x + 4 + \frac{2}{3} arctan (\frac{1}{3} (2 x + 1)) - 23 arctan (\frac{1}{3} (2 x + 1))

= 4 x + 2 ln ∣ x - 1 ∣ - ln 4 x^{2} + 4 x + 4 + \frac{2}{3} arctan (\frac{1}{3} (2 x + 1)) - 23 arctan (\frac{1}{3} (2 x + 1))

Füge eine Konstante zur Lösung hinzu

= 4 x + 2 ln ∣ x - 1 ∣ - ln 4 x^{2} + 4 x + 4 + \frac{2}{3} arctan (\frac{1}{3} (2 x + 1)) - 23 arctan (\frac{1}{3} (2 x + 1)) + C

\int d arcsin (x) d x

\int 7 sin^{7} (x) d x

\int sin (\frac{π x}{l}) d x

\int cos (\frac{2 π x}{L}) d x

\int 2 x^{2} d x