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

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

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

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

Solution steps

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

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{4 x - 2}{( x - 1 ) ^{2}} d x

Apply Integration By Parts

: - \frac{4 x - 2}{x - 1} - \int - \frac{4}{x - 1} d x

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

\int - \frac{4}{x - 1} d x = - 4 ln ∣ x - 1 ∣

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

Simplify

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

Add a constant to the solution

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

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

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

\int \frac{cos ( 2 θ )}{sin ^{2} ( 2 θ )} d θ

\int 500 x^{\frac{1}{4}} d x

\int e^{- inπ t} d t

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