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

\frac{\partial}{\partial x} (x y + x + y)

(y^{- 1})^{^{'}}

x \to - 1 lim (\frac{x + 1}{x ^{3} + 1})

y^{^{'}} = - y + \frac{1}{y}

2 - 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