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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

= \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

\frac{1}{s ( s + 1 ) ^{3}}

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

y = 8 x + 3 x^{\frac{1}{6}}

\int_{0}^{1} π (4 - \frac{1}{2} x)^{2} d x

(2 x - 9) (x^{2} + 6), at x = 3

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