integral of ((y^2-14y+5))/(y(y-1)(y-5))

Lösung

\int \frac{( y ^{2} - 14 y + 5 )}{y ( y - 1 ) ( y - 5 )} d y

\frac{1}{5} ln ∣ y ∣ + \frac{4}{5} ln ∣ y - 5 ∣ - 12 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ y - 5 ∣) - 8 (\frac{1}{5} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 5 ∣) + C

Schritte zur Lösung

\int \frac{y ^{2} - 14 y + 5}{y ( y - 1 ) ( y - 5 )} d y

Wende U-Substitution an

= \int \frac{u ^{2} - 12 u - 8}{u ( u + 1 ) ( u - 4 )} d u

Multipliziere aus

\frac{u ^{2} - 12 u - 8}{u ( u + 1 ) ( u - 4 )} : \frac{u ^{2}}{u ( u + 1 ) ( u - 4 )} - \frac{12 u}{u ( u + 1 ) ( u - 4 )} - \frac{8}{u ( u + 1 ) ( u - 4 )}

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{u ^{2}}{u ( u + 1 ) ( u - 4 )} d u - \int \frac{12 u}{u ( u + 1 ) ( u - 4 )} d u - \int \frac{8}{u ( u + 1 ) ( u - 4 )} d u

\int \frac{u ^{2}}{u ( u + 1 ) ( u - 4 )} d u = \frac{1}{5} ln ∣ u + 1 ∣ + \frac{4}{5} ln ∣ u - 4 ∣

\int \frac{12 u}{u ( u + 1 ) ( u - 4 )} d u = 12 (- \frac{1}{5} ln ∣ u + 1 ∣ + \frac{1}{5} ln ∣ u - 4 ∣)

\int \frac{8}{u ( u + 1 ) ( u - 4 )} d u = 8 (\frac{1}{5} ln ∣ u + 1 ∣ - \frac{1}{4} ln ∣ u ∣ + \frac{1}{20} ln ∣ u - 4 ∣)

= \frac{1}{5} ln ∣ u + 1 ∣ + \frac{4}{5} ln ∣ u - 4 ∣ - 12 (- \frac{1}{5} ln ∣ u + 1 ∣ + \frac{1}{5} ln ∣ u - 4 ∣) - 8 (\frac{1}{5} ln ∣ u + 1 ∣ - \frac{1}{4} ln ∣ u ∣ + \frac{1}{20} ln ∣ u - 4 ∣)

Setze in

u = y - 1

ein

= \frac{1}{5} ln ∣ y - 1 + 1 ∣ + \frac{4}{5} ln ∣ y - 1 - 4 ∣ - 12 (- \frac{1}{5} ln ∣ y - 1 + 1 ∣ + \frac{1}{5} ln ∣ y - 1 - 4 ∣) - 8 (\frac{1}{5} ln ∣ y - 1 + 1 ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 1 - 4 ∣)

Vereinfache

\frac{1}{5} ln ∣ y - 1 + 1 ∣ + \frac{4}{5} ln ∣ y - 1 - 4 ∣ - 12 (- \frac{1}{5} ln ∣ y - 1 + 1 ∣ + \frac{1}{5} ln ∣ y - 1 - 4 ∣) - 8 (\frac{1}{5} ln ∣ y - 1 + 1 ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 1 - 4 ∣) : \frac{1}{5} ln ∣ y ∣ + \frac{4}{5} ln ∣ y - 5 ∣ - 12 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ y - 5 ∣) - 8 (\frac{1}{5} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 5 ∣)

= \frac{1}{5} ln ∣ y ∣ + \frac{4}{5} ln ∣ y - 5 ∣ - 12 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ y - 5 ∣) - 8 (\frac{1}{5} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 5 ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} ln ∣ y ∣ + \frac{4}{5} ln ∣ y - 5 ∣ - 12 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ y - 5 ∣) - 8 (\frac{1}{5} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 5 ∣) + C

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

\int 7 x + 14 x ln (3 x) d x

\int x \cdot sin^{2} (2 x) d x

\int \frac{x}{x ^{2} + 8 x + 32} d x

\int x - e^{x} d x