What is the integral of (5-4y^2)/(3+2y) ?

The integral of (5-4y^2)/(3+2y) is -y^2+3y-2ln|2y+3|+27/4+C

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integral of (5-4y^2)/(3+2y)

\int \frac{5 - 4 y ^{2}}{3 + 2 y} d y

- y^{2} + 3 y - 2 ln ∣ 2 y + 3 ∣ + \frac{27}{4} + C

Solution steps

\int \frac{5 - 4 y ^{2}}{3 + 2 y} d y

Expand

\frac{5 - 4 y ^{2}}{3 + 2 y} : \frac{5}{3 + 2 y} - \frac{4 y ^{2}}{3 + 2 y}

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{5}{3 + 2 y} d y - \int \frac{4 y ^{2}}{3 + 2 y} d y

\int \frac{5}{3 + 2 y} d y = \frac{5}{2} ln ∣ 3 + 2 y ∣

\int \frac{4 y ^{2}}{3 + 2 y} d y = \frac{1}{4} (4 y^{2} - 12 y + 18 ln ∣ 2 y + 3 ∣ - 27)

= \frac{5}{2} ln ∣ 3 + 2 y ∣ - \frac{1}{4} (4 y^{2} - 12 y + 18 ln ∣ 2 y + 3 ∣ - 27)

Simplify

\frac{5}{2} ln ∣ 3 + 2 y ∣ - \frac{1}{4} (4 y^{2} - 12 y + 18 ln ∣ 2 y + 3 ∣ - 27) : - y^{2} + 3 y - 2 ln ∣ 2 y + 3 ∣ + \frac{27}{4}

= - y^{2} + 3 y - 2 ln ∣ 2 y + 3 ∣ + \frac{27}{4}

Add a constant to the solution

= - y^{2} + 3 y - 2 ln ∣ 2 y + 3 ∣ + \frac{27}{4} + C

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What is the integral of (5-4y^2)/(3+2y) ?
The integral of (5-4y^2)/(3+2y) is -y^2+3y-2ln|2y+3|+27/4+C