integral of (6x^2-3x+5)/((2x-1)(x^2+1))

Lösung

\int \frac{6 x ^{2} - 3 x + 5}{( 2 x - 1 ) ( x ^{2} + 1 )} d x

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

Schritte zur Lösung

\int \frac{6 x ^{2} - 3 x + 5}{( 2 x - 1 ) ( x ^{2} + 1 )} d x

Ermittle den Partialbruch von

\frac{6 x ^{2} - 3 x + 5}{( 2 x - 1 ) ( x ^{2} + 1 )} : \frac{4}{2 x - 1} + \frac{x - 1}{x ^{2} + 1}

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

\int \frac{4}{2 x - 1} d x = 2 ln ∣ 2 x - 1 ∣

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{2 x ^{2}}{x + 6} d x

\int sin^{2} (x) + 3 d x

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

\int - \frac{2}{81 + 25 t ^{2}} d t

\int \frac{sin ( 2 x )}{( 1 - cos ( 2 x ) )} d x