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

Lösung

integral

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

2 x + \frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} - \frac{1}{2} (x - 1)^{2} - 2 + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l (2 x - 1) e^{x}

in t e g r a l \frac{( x ^{2} + 2 )}{( x + 1 )}

in t e g r a l 9 cos^{2} (x)

in t e g r a l x lo g_{10} (x^{2} + 5)

f (x) = (x^{2} + x + 1)^{\frac{1}{2}}