What is the integral of ((x^2-3x-7))/((2x+3)(x+1)^2) ?

The integral of ((x^2-3x-7))/((2x+3)(x+1)^2) is -1/2 ln|2x+3|+ln|x+1|+3/(x+1)+C

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integral of ((x^2-3x-7))/((2x+3)(x+1)^2)

\int \frac{( x ^{2} - 3 x - 7 )}{( 2 x + 3 ) ( x + 1 ) ^{2}} d x

- \frac{1}{2} ln ∣ 2 x + 3 ∣ + ln ∣ x + 1 ∣ + \frac{3}{x + 1} + C

Solution steps

\int \frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 ) ^{2}} : - \frac{1}{2 x + 3} + \frac{1}{x + 1} - \frac{3}{( x + 1 ) ^{2}}

= \int - \frac{1}{2 x + 3} + \frac{1}{x + 1} - \frac{3}{( x + 1 ) ^{2}} d 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{1}{2 x + 3} d x + \int \frac{1}{x + 1} d x - \int \frac{3}{( x + 1 ) ^{2}} d x

\int \frac{1}{2 x + 3} d x = \frac{1}{2} ln ∣ 2 x + 3 ∣

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

\int \frac{3}{( x + 1 ) ^{2}} d x = - \frac{3}{x + 1}

= - \frac{1}{2} ln ∣ 2 x + 3 ∣ + ln ∣ x + 1 ∣ - (- \frac{3}{x + 1})

Simplify

= - \frac{1}{2} ln ∣ 2 x + 3 ∣ + ln ∣ x + 1 ∣ + \frac{3}{x + 1}

Add a constant to the solution

= - \frac{1}{2} ln ∣ 2 x + 3 ∣ + ln ∣ x + 1 ∣ + \frac{3}{x + 1} + C

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

t^{3} d x + (x t^{2} + x^{3}) d t = 0

\frac{d P}{d t} = (\frac{9}{240}) P - \frac{1}{4800} (P)^{2}, P (0) = 240

3 x^{4} - 6 x^{3} + 1

f (x) = \frac{x ^{2} + 8}{9 x - 2}

What is the integral of ((x^2-3x-7))/((2x+3)(x+1)^2) ?
The integral of ((x^2-3x-7))/((2x+3)(x+1)^2) is -1/2 ln|2x+3|+ln|x+1|+3/(x+1)+C