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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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{d}{d x} (\frac{x ^{2}}{x ^{4} + 1})

\frac{d y}{d x} = 2 y - 1

x^{2} \frac{d y}{d x} + y^{2} = x y

\int cot^{2} (x) csc (x) d x

\frac{\partial}{\partial x} (- e^{2 y} cos (2 x) \cdot 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