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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Apply u-substitution

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

Expand

\frac{u + 2}{u ( u + 1 ) ^{2}} : \frac{u}{u ( u + 1 ) ^{2}} + \frac{2}{u ( u + 1 ) ^{2}}

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

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

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

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

Substitute back

u = x - 2

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{( 2 x + 2 )}{x ^{2} + 10 x + 24} d x

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

\int ln (x) 3 x d x

\int 5 (z - 4) 3 z^{2} - 8 z d z

\int \frac{8 s ^{3}}{5 - s ^{4}} d s

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