integral of (6x^2+3x+10)/(x^3+2x^2+5x)

Lösung

\int \frac{6 x ^{2} + 3 x + 10}{x ^{3} + 2 x ^{2} + 5 x} d x

- \frac{5}{2} arctan (\frac{1}{2} (x + 1)) + 2 ln x^{2} + 2 x + 5 + 2 ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{6 x ^{2} + 3 x + 10}{x ^{3} + 2 x ^{2} + 5 x} d x

Multipliziere aus

\frac{6 x ^{2} + 3 x + 10}{x ^{3} + 2 x ^{2} + 5 x} : \frac{6 x ^{2}}{x ^{3} + 2 x ^{2} + 5 x} + \frac{3 x}{x ^{3} + 2 x ^{2} + 5 x} + \frac{10}{x ^{3} + 2 x ^{2} + 5 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{6 x ^{2}}{x ^{3} + 2 x ^{2} + 5 x} d x + \int \frac{3 x}{x ^{3} + 2 x ^{2} + 5 x} d x + \int \frac{10}{x ^{3} + 2 x ^{2} + 5 x} d x

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

\int \frac{3 x}{x ^{3} + 2 x ^{2} + 5 x} d x = \frac{3}{2} arctan (\frac{1}{2} (x + 1))

\int \frac{10}{x ^{3} + 2 x ^{2} + 5 x} d x = 2 ln ∣ x ∣ - ln x^{2} + 2 x + 5 - arctan (\frac{1}{2} (x + 1))

= 6 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1))) + \frac{3}{2} arctan (\frac{1}{2} (x + 1)) + 2 ln ∣ x ∣ - ln x^{2} + 2 x + 5 - arctan (\frac{1}{2} (x + 1))

Vereinfache

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

= - \frac{5}{2} arctan (\frac{1}{2} (x + 1)) + 2 ln x^{2} + 2 x + 5 + 2 ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{5}{2} arctan (\frac{1}{2} (x + 1)) + 2 ln x^{2} + 2 x + 5 + 2 ln ∣ x ∣ + C

t an g e n t x^{2} + 40

\frac{d}{d x} (\frac{x ^{5} - 2 x}{- 5})

d er i v a t i v e 7 u^{e}

ma c l a u r in e^{- x^{2}}

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