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

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

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

2 ln x + x - 1 + C

Solution steps

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

Apply u-substitution

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

(u + 1) u = (u + 1) u

= \int \frac{1}{u + 1 u} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= 2 ln v^{2} + 1 + v

Substitute back

= 2 ln (x - 1)^{2} + 1 + x - 1

(x - 1)^{2} + 1 = x

= 2 ln x + x - 1

Add a constant to the solution

= 2 ln x + x - 1 + C

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

\int x in x d x

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

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

\int \frac{2 x ^{2} + x}{( 8 x ^{3} + 6 x ^{2} ) ^{2}} d x

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