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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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

\int 128 + 128 cos (x) d x

\int \frac{3}{x - 5} d x

\int \frac{1}{x ^{2} 36 x ^{2} - 25} d x

in t e g r a l x^{- 5}