integral of (7x^3+7x)/(x-1)

Lösung

\int \frac{7 x ^{3} + 7 x}{x - 1} d x

28 x + 14 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 28 + C

Schritte zur Lösung

\int \frac{7 x ^{3} + 7 x}{x - 1} d x

Multipliziere aus

\frac{7 x ^{3} + 7 x}{x - 1} : \frac{7 x ^{3}}{x - 1} + \frac{7 x}{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{7 x ^{3}}{x - 1} d x + \int \frac{7 x}{x - 1} d x

\int \frac{7 x ^{3}}{x - 1} d x = 7 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣)

\int \frac{7 x}{x - 1} d x = 7 (x - 1 + ln ∣ x - 1 ∣)

= 7 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣) + 7 (x - 1 + ln ∣ x - 1 ∣)

Vereinfache

7 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣) + 7 (x - 1 + ln ∣ x - 1 ∣) : 28 x + 14 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 28

= 28 x + 14 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 28

Füge eine Konstante zur Lösung hinzu

= 28 x + 14 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 28 + C

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