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

The integral of (3x-4)/((x-1)(x+2)) is 3ln|x-1|-10(-1/3 ln|x+2|+1/3 ln|x-1|)+C

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

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

3 ln ∣ x - 1 ∣ - 10 (- \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 1 ∣) + C

Solution steps

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

Apply u-substitution

= \int \frac{3 u - 10}{u ( u - 3 )} d u

Expand

\frac{3 u - 10}{u ( u - 3 )} : \frac{3 u}{u ( u - 3 )} - \frac{10}{u ( u - 3 )}

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}{u ( u - 3 )} d u - \int \frac{10}{u ( u - 3 )} d u

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

\int \frac{10}{u ( u - 3 )} d u = 10 (- \frac{1}{3} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 3 ∣)

= 3 ln ∣ u - 3 ∣ - 10 (- \frac{1}{3} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 3 ∣)

Substitute back

u = x + 2

= 3 ln ∣ x + 2 - 3 ∣ - 10 (- \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x + 2 - 3 ∣)

Simplify

3 ln ∣ x + 2 - 3 ∣ - 10 (- \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x + 2 - 3 ∣) : 3 ln ∣ x - 1 ∣ - 10 (- \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 1 ∣)

= 3 ln ∣ x - 1 ∣ - 10 (- \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 1 ∣)

Add a constant to the solution

= 3 ln ∣ x - 1 ∣ - 10 (- \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 1 ∣) + C

x^{^{''}} + 2 x^{^{'}} + 5 x = 0, x (0) = 1, x^{^{'}} (0) = - 1

x \to 2 lim (3 x^{2} - 2 x + 4)

\int (x^{3} + x)^{8} (3 x^{2} + 1) d x

\int \frac{6}{2 x + 3 x ^{2}} d x

\frac{d}{d x} (\frac{e ^{x}}{9 e ^{x} + x ^{2}})

What is the integral of (3x-4)/((x-1)(x+2)) ?
The integral of (3x-4)/((x-1)(x+2)) is 3ln|x-1|-10(-1/3 ln|x+2|+1/3 ln|x-1|)+C