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

The integral of 1/((x^2-1)(x+2)) is 1/3 ln|x+2|-1/2 ln|x+1|+1/6 ln|x-1|+C

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

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

\frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( u ^{2} - 4 u + 3 )} d u

= \int \frac{1}{u ( u ^{2} - 4 u + 3 )} d u

Take the partial fraction of

\frac{1}{u ( u ^{2} - 4 u + 3 )} : \frac{1}{3 u} - \frac{1}{2 ( u - 1 )} + \frac{1}{6 ( u - 3 )}

= \int \frac{1}{3 u} - \frac{1}{2 ( u - 1 )} + \frac{1}{6 ( u - 3 )} d u

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}{3 u} d u - \int \frac{1}{2 ( u - 1 )} d u + \int \frac{1}{6 ( u - 3 )} d u

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{2 ( u - 1 )} d u = \frac{1}{2} ln ∣ u - 1 ∣

\int \frac{1}{6 ( u - 3 )} d u = \frac{1}{6} ln ∣ u - 3 ∣

= \frac{1}{3} ln ∣ u ∣ - \frac{1}{2} ln ∣ u - 1 ∣ + \frac{1}{6} ln ∣ u - 3 ∣

Substitute back

u = x + 2

= \frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 2 - 1 ∣ + \frac{1}{6} ln ∣ x + 2 - 3 ∣

Simplify

\frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 2 - 1 ∣ + \frac{1}{6} ln ∣ x + 2 - 3 ∣ : \frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 1 ∣

= \frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 1 ∣ + C

\int (k e^{- k x}) d x

y^{^{'}} + 3 (tan (3 x)) y = cos (3 x)

\frac{( 12 s + 5 )}{s ( s + 3 )}

\frac{1}{x ^{2}} - \frac{7}{x ^{4}}

\int (2 x - 3) e^{x^{2} - 3 x} d x

What is the integral of 1/((x^2-1)(x+2)) ?
The integral of 1/((x^2-1)(x+2)) is 1/3 ln|x+2|-1/2 ln|x+1|+1/6 ln|x-1|+C