integral of (8x^2+x+9)/((x+1)(x^2+1))

Lösung

\int \frac{8 x ^{2} + x + 9}{( x + 1 ) ( x ^{2} + 1 )} d x

8 ln ∣ x + 1 ∣ + arctan (x) + C

Schritte zur Lösung

\int \frac{8 x ^{2} + x + 9}{( x + 1 ) ( x ^{2} + 1 )} d x

Multipliziere aus

\frac{8 x ^{2} + x + 9}{( x + 1 ) ( x ^{2} + 1 )} : \frac{8 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} + \frac{x}{( x + 1 ) ( x ^{2} + 1 )} + \frac{9}{( x + 1 ) ( x ^{2} + 1 )}

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{8 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x + \int \frac{x}{( x + 1 ) ( x ^{2} + 1 )} d x + \int \frac{9}{( x + 1 ) ( x ^{2} + 1 )} d x

\int \frac{8 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x = 8 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x)))

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

\int \frac{9}{( x + 1 ) ( x ^{2} + 1 )} d x = 9 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)))

= 8 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x))) - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) + 9 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)))

Vereinfache

8 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x))) - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) + 9 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x))) : 8 ln ∣ x + 1 ∣ + arctan (x)

= 8 ln ∣ x + 1 ∣ + arctan (x)

Füge eine Konstante zur Lösung hinzu

= 8 ln ∣ x + 1 ∣ + arctan (x) + C

\int \frac{9}{3 e ^{- 2 v} - 7} d v

\int \frac{e ^{2 x}}{1 + 6 e ^{2 x} + 9 e ^{4 x}} d x

\int sin (7 x) cos (8 x) d x

\int 5 sin (3 x) - 4 x sec (x^{2}) d x

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