What is the integral of 1/(x(x-0.5)) ?

The integral of 1/(x(x-0.5)) is 2(-ln(2|x|)+ln|2x-1|)+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(x-0.5))

\int \frac{1}{x ( x - 0.5 )} d x

2 (- ln (2 ∣ x ∣) + ln ∣ 2 x - 1 ∣) + C

Solution steps

\int \frac{1}{x ( x - 0.5 )} d x

x - 0.5 = 0.5 (\frac{x}{0.5} - 1)

= \int \frac{1}{x \cdot 0.5 ( \frac{x}{0.5} - 1 )} d x

Take the constant out:

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

= \frac{1}{0.5} \cdot \int \frac{1}{x ( \frac{x}{0.5} - 1 )} d x

Simplify

= 2 \cdot \int \frac{1}{x ( \frac{x}{0.5} - 1 )} d x

Apply u-substitution

: \int \frac{1}{u ( u - 1 )} d u

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

Take the partial fraction of

\frac{1}{u ( u - 1 )} : - \frac{1}{u} + \frac{1}{u - 1}

= 2 \cdot \int - \frac{1}{u} + \frac{1}{u - 1} 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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= 2 (- ln ∣ u ∣ + ln ∣ u - 1 ∣)

Substitute back

u = \frac{x}{0.5}

= 2 (- ln \frac{x}{0.5} + ln \frac{x}{0.5} - 1)

Simplify

2 (- ln \frac{x}{0.5} + ln \frac{x}{0.5} - 1) : 2 (- ln (2 ∣ x ∣) + ln ∣ 2 x - 1 ∣)

= 2 (- ln (2 ∣ x ∣) + ln ∣ 2 x - 1 ∣)

Add a constant to the solution

= 2 (- ln (2 ∣ x ∣) + ln ∣ 2 x - 1 ∣) + C

\frac{d}{d x} (x^{2} - 9 x)

\int_{0}^{\frac{π}{2}} 3 cos^{2} (x) d x

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

\int 2^{3 x + 4} d x

x \to 0 lim (cos (7 x))

What is the integral of 1/(x(x-0.5)) ?
The integral of 1/(x(x-0.5)) is 2(-ln(2|x|)+ln|2x-1|)+C