What is the integral of (16y-9)/(3y-5y^2) ?

The integral of (16y-9)/(3y-5y^2) is -16/5 ln|3-5y|+9(-1/3 ln|y|+1/3 ln|5y-3|)+C

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integral of (16y-9)/(3y-5y^2)

\int \frac{16 y - 9}{3 y - 5 y ^{2}} d y

- \frac{16}{5} ln ∣ 3 - 5 y ∣ + 9 (- \frac{1}{3} ln ∣ y ∣ + \frac{1}{3} ln ∣ 5 y - 3 ∣) + C

Solution steps

\int \frac{16 y - 9}{3 y - 5 y ^{2}} d y

Expand

\frac{16 y - 9}{3 y - 5 y ^{2}} : \frac{16 y}{3 y - 5 y ^{2}} - \frac{9}{3 y - 5 y ^{2}}

Apply the Sum Rule:

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

= \int \frac{16 y}{3 y - 5 y ^{2}} d y - \int \frac{9}{3 y - 5 y ^{2}} d y

\int \frac{16 y}{3 y - 5 y ^{2}} d y = - \frac{16}{5} ln ∣ 3 - 5 y ∣

\int \frac{9}{3 y - 5 y ^{2}} d y = - 9 (- \frac{1}{3} ln ∣ y ∣ + \frac{1}{3} ln ∣ 5 y - 3 ∣)

= - \frac{16}{5} ln ∣ 3 - 5 y ∣ - (- 9 (- \frac{1}{3} ln ∣ y ∣ + \frac{1}{3} ln ∣ 5 y - 3 ∣))

Simplify

= - \frac{16}{5} ln ∣ 3 - 5 y ∣ + 9 (- \frac{1}{3} ln ∣ y ∣ + \frac{1}{3} ln ∣ 5 y - 3 ∣)

Add a constant to the solution

= - \frac{16}{5} ln ∣ 3 - 5 y ∣ + 9 (- \frac{1}{3} ln ∣ y ∣ + \frac{1}{3} ln ∣ 5 y - 3 ∣) + C

\int x^{1} cos (- 9 x) d x

x \to 3 lim (\frac{∣ x - 2 ∣}{x - 2})

y = \frac{4 x ^{7}}{7}

\frac{d}{d x} (e^{3 x - 5})

\int arcsin (x) d x

What is the integral of (16y-9)/(3y-5y^2) ?
The integral of (16y-9)/(3y-5y^2) is -16/5 ln|3-5y|+9(-1/3 ln|y|+1/3 ln|5y-3|)+C