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

Lösung

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

\frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - x + 2 ln ∣ x + 1 ∣ + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - x + 2 ln ∣ x + 1 ∣ + C

\int \frac{1}{( x - 1 ) ( x - 4 )} d x

t an g e n t f (x) = \frac{1}{3 + 2 x}, at x = 2

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

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

\frac{d y}{d x} = e^{3 x + 2 y}