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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

y = \frac{e ^{x} + e ^{- x}}{x}

\frac{d}{d x} (\frac{500}{1 + 0.06 x ^{2}})

\int \frac{1}{x + 4} d x

\frac{\partial}{\partial y} (3 (x - y) e^{3 y + 5 x^{2}})

\int \frac{9 - x ^{2}}{x} d x

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