What is the integral of (3x^2+5x-4)/((x^2-1)(x+1)) ?

The integral of (3x^2+5x-4)/((x^2-1)(x+1)) is 2ln|x+1|+3ln|x-1|-1/2 ln|x-1|-3/(x+1)-3/2 ln|x-1|+C

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integral of (3x^2+5x-4)/((x^2-1)(x+1))

\int \frac{3 x ^{2} + 5 x - 4}{( x ^{2} - 1 ) ( x + 1 )} d x

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

Solution steps

\int \frac{3 x ^{2} + 5 x - 4}{( x ^{2} - 1 ) ( x + 1 )} d x

Apply u-substitution

: \int \frac{3 u ^{2} - u - 6}{u ( u ^{2} - 2 u )} d u

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

Expand

\frac{3 u ^{2} - u - 6}{u ( u ^{2} - 2 u )} : \frac{3 u ^{2}}{u ( u ^{2} - 2 u )} - \frac{u}{u ( u ^{2} - 2 u )} - \frac{6}{u ( u ^{2} - 2 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{3 u ^{2}}{u ( u ^{2} - 2 u )} d u - \int \frac{u}{u ( u ^{2} - 2 u )} d u - \int \frac{6}{u ( u ^{2} - 2 u )} d u

\int \frac{3 u ^{2}}{u ( u ^{2} - 2 u )} d u = 3 ln ∣ u - 2 ∣

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

\int \frac{6}{u ( u ^{2} - 2 u )} d u = 6 (- \frac{1}{4} ln ∣ u ∣ + \frac{1}{2 u} + \frac{1}{4} ln ∣ u - 2 ∣)

= 3 ln ∣ u - 2 ∣ - (- \frac{1}{2} ln ∣ u ∣ + \frac{1}{2} ln ∣ u - 2 ∣) - 6 (- \frac{1}{4} ln ∣ u ∣ + \frac{1}{2 u} + \frac{1}{4} ln ∣ u - 2 ∣)

Substitute back

u = x + 1

= 3 ln ∣ x + 1 - 2 ∣ - (- \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 1 - 2 ∣) - 6 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x + 1 - 2 ∣)

Simplify

3 ln ∣ x + 1 - 2 ∣ - (- \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 1 - 2 ∣) - 6 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x + 1 - 2 ∣) : 2 ln ∣ x + 1 ∣ + 3 ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{3}{x + 1} - \frac{3}{2} ln ∣ x - 1 ∣

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (- 4 sin (x))

\frac{d}{d x} (((- 3 x + 1)^{4} - 5)^{3})

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

2 - 2 x^{2}

What is the integral of (3x^2+5x-4)/((x^2-1)(x+1)) ?
The integral of (3x^2+5x-4)/((x^2-1)(x+1)) is 2ln|x+1|+3ln|x-1|-1/2 ln|x-1|-3/(x+1)-3/2 ln|x-1|+C