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

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

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integral of 2/(x^3+3x^2+2x)

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

ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{x ^{3} + 3 x ^{2} + 2 x} d x

Take the partial fraction of

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

= 2 \cdot \int \frac{1}{2 x} - \frac{1}{x + 1} + \frac{1}{2 ( x + 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

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

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

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

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

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

Simplify

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

= ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣

Add a constant to the solution

= ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣ + C

\int \frac{1}{a ^{2} u ^{2}} d u

f (x) = x^{2} - 4, (- 3, 5)

\frac{d y}{d t} = 1.5 y, y (0) = 100

\int_{0}^{\infty} t e^{- b t} d t

f (x) = - 8 x^{4} + 11 x^{2} - 3, (1, 0)

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