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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{\partial}{\partial y} (3 x e^{- 2 x y})

\int \frac{1}{x ^{2} + 4 x + 8} d x

5 e^{2 x} \frac{d y}{d x} = - 10 \frac{x}{y ^{2}}

t an g e n t f (x) = \frac{1}{2 x}, at x = 49

d er i v a t i v e f (x) = 9 - \frac{4}{x ^{2}}