What is the integral of (y^3)/(2y-1) ?

The integral of (y^3)/(2y-1) is 1/96 (2(2y-1)^3+9(2y-1)^2+18(2y-1)+6ln|2y-1|)+C

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

\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

Solution steps

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

Apply u-substitution

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

Take the constant out:

\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

Expand

\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

Apply the Sum Rule:

\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 ∣)

Substitute back

u = 2 y - 1

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

Simplify

\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 ∣)

Add a constant to the solution

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

What is the integral of (y^3)/(2y-1) ?
The integral of (y^3)/(2y-1) is 1/96 (2(2y-1)^3+9(2y-1)^2+18(2y-1)+6ln|2y-1|)+C