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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln ∣ 4 x - 2 ∣ - ln ∣ 4 x - 4 ∣ + C

Solution steps

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

Complete the square

(x - 1) (1 - 2 x) : - 2 (x - \frac{3}{4})^{2} + \frac{1}{8}

= \int \frac{1}{- 2 ( x - \frac{3}{4} ) ^{2} + \frac{1}{8}} d x

Apply u-substitution

: \int \frac{8}{- 16 u ^{2} + 1} d u

= \int \frac{8}{- 16 u ^{2} + 1} d u

Take the constant out:

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

= 8 \cdot \int \frac{1}{- 16 u ^{2} + 1} d u

Apply Integral Substitution

= 8 \cdot \int \frac{1}{4 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 8 \cdot \frac{1}{4} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= 8 \cdot \frac{1}{4} (\frac{ln 4 ( x - \frac{3}{4} ) + 1}{2} - \frac{ln 4 ( x - \frac{3}{4} ) - 1}{2})

Simplify

8 \cdot \frac{1}{4} (\frac{ln 4 ( x - \frac{3}{4} ) + 1}{2} - \frac{ln 4 ( x - \frac{3}{4} ) - 1}{2}) : ln ∣ 4 x - 2 ∣ - ln ∣ 4 x - 4 ∣

= ln ∣ 4 x - 2 ∣ - ln ∣ 4 x - 4 ∣

Add a constant to the solution

= ln ∣ 4 x - 2 ∣ - ln ∣ 4 x - 4 ∣ + C

\frac{d}{d x} (cos (f (x)))

2 y \frac{d y}{d x} - x y^{2} = x

\frac{d}{d x} (arcsin (x) - 2 x)

\frac{( 5 x ^{2} + 9 ) ^{5}}{( x - 3 ) ^{4}}

\int x y^{3} d x

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