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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int \frac{u + 1}{u ^{2}} d u

= 3 \cdot \int \frac{u + 1}{u ^{2}} d u

Expand

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

= 3 \cdot \int \frac{1}{u} + \frac{1}{u ^{2}} 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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 3 (ln ∣ u ∣ - \frac{1}{u})

Substitute back

u = x - 1

= 3 (ln ∣ x - 1 ∣ - \frac{1}{x - 1})

Add a constant to the solution

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

\int (\frac{4}{x ^{3}} + 6 x) d x

\int x \cdot sin (6 x) d x

\int \frac{e ^{x}}{e ^{2 x} + 2 e ^{x} + 2} d x

\int 2 x (6 - x) d x

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

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