integral of (y^3)/(2y-1)

Lösung

\int \frac{y ^{3}}{2 y - 1} d y

\frac{1}{96} (2 (2 y - 1)^{3} + 9 (2 y - 1)^{2} + 18 (2 y - 1) + 6 ln ∣ 2 y - 1 ∣) + C

Schritte zur Lösung

\int \frac{y ^{3}}{2 y - 1} d y

Wende U-Substitution an

= \int \frac{( u + 1 ) ^{3}}{16 u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{16} \cdot \int \frac{( u + 1 ) ^{3}}{u} d u

Multipliziere aus

\frac{( u + 1 ) ^{3}}{u} : u^{2} + 3 u + 3 + \frac{1}{u}

= \frac{1}{16} \cdot \int u^{2} + 3 u + 3 + \frac{1}{u} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{16} (\int u^{2} d u + \int 3 u d u + \int 3 d u + \int \frac{1}{u} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 3 u d u = \frac{3 u ^{2}}{2}

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{16} (\frac{u ^{3}}{3} + \frac{3 u ^{2}}{2} + 3 u + ln ∣ u ∣)

Setze in

u = 2 y - 1

ein

= \frac{1}{16} (\frac{( 2 y - 1 ) ^{3}}{3} + \frac{3 ( 2 y - 1 ) ^{2}}{2} + 3 (2 y - 1) + ln ∣ 2 y - 1 ∣)

Vereinfache

\frac{1}{16} (\frac{( 2 y - 1 ) ^{3}}{3} + \frac{3 ( 2 y - 1 ) ^{2}}{2} + 3 (2 y - 1) + ln ∣ 2 y - 1 ∣) : \frac{1}{96} (2 (2 y - 1)^{3} + 9 (2 y - 1)^{2} + 18 (2 y - 1) + 6 ln ∣ 2 y - 1 ∣)

= \frac{1}{96} (2 (2 y - 1)^{3} + 9 (2 y - 1)^{2} + 18 (2 y - 1) + 6 ln ∣ 2 y - 1 ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{96} (2 (2 y - 1)^{3} + 9 (2 y - 1)^{2} + 18 (2 y - 1) + 6 ln ∣ 2 y - 1 ∣) + C

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

\int 2 sin (4 x) ln (sin (4 x)) d x

\int \frac{6 ( x ^{2} + 4 )}{x ( x ^{2} + 6 )} d x

\int \frac{1}{y ^{2} + 6 y + 10} d y

\int \frac{x + 1}{x ^{3} - x ^{2} - 6 x} d x