What is the integral of (6x^2+13x+6)/((x+2)(x+1)^2) ?

The integral of (6x^2+13x+6)/((x+2)(x+1)^2) is 4ln|x+2|+2ln|x+1|+1/(x+1)+C

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integral of (6x^2+13x+6)/((x+2)(x+1)^2)

\int \frac{6 x ^{2} + 13 x + 6}{( x + 2 ) ( x + 1 ) ^{2}} d x

4 ln ∣ x + 2 ∣ + 2 ln ∣ x + 1 ∣ + \frac{1}{x + 1} + C

Solution steps

\int \frac{6 x ^{2} + 13 x + 6}{( x + 2 ) ( x + 1 ) ^{2}} d x

Take the partial fraction of

\frac{6 x ^{2} + 13 x + 6}{( x + 2 ) ( x + 1 ) ^{2}} : \frac{4}{x + 2} + \frac{2}{x + 1} - \frac{1}{( x + 1 ) ^{2}}

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

\int \frac{4}{x + 2} d x = 4 ln ∣ x + 2 ∣

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

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

= 4 ln ∣ x + 2 ∣ + 2 ln ∣ x + 1 ∣ - (- \frac{1}{x + 1})

Simplify

= 4 ln ∣ x + 2 ∣ + 2 ln ∣ x + 1 ∣ + \frac{1}{x + 1}

Add a constant to the solution

= 4 ln ∣ x + 2 ∣ + 2 ln ∣ x + 1 ∣ + \frac{1}{x + 1} + C

y = e^{x^{3}}, (- 1, \frac{1}{e})

\frac{( y - 2 ) ^{3}}{( y ^{2} + 4 y ) ^{7}}

x \to + 3 lim (f (x))

x \to 90 lim (sin (x))

\frac{d y}{d x} = \frac{x}{16 y}

What is the integral of (6x^2+13x+6)/((x+2)(x+1)^2) ?
The integral of (6x^2+13x+6)/((x+2)(x+1)^2) is 4ln|x+2|+2ln|x+1|+1/(x+1)+C