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

The integral of 1/(sqrt((x-1)(2-x))) is arcsin(2x-3)+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-1)(2-x)))

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

arcsin (2 x - 3) + C

Solution steps

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

Apply u-substitution

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

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

Expand

u (- u + 1) : - u^{2} + u

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

Complete the square

- u^{2} + u : - (u - \frac{1}{2})^{2} + \frac{1}{4}

= \int \frac{1}{- ( u - \frac{1}{2} ) ^{2} + \frac{1}{4}} d u

Apply u-substitution

: \int \frac{2}{- 4 v ^{2} + 1} d v

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d w

= 2 \cdot \int \frac{1}{2} d w

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot 1 \cdot w

Substitute back

= 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 (x - 1 - \frac{1}{2}))

Simplify

2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 (x - 1 - \frac{1}{2})) : arcsin (2 x - 3)

= arcsin (2 x - 3)

Add a constant to the solution

= arcsin (2 x - 3) + C

\int_{- 3}^{3} x^{2} - 3 d x

y = x^{3} - 3 x

\frac{d}{d x} (2 \cdot x \cdot y)

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

\frac{d}{d t} (a cos (2 t) + b sin (2 t))

What is the integral of 1/(sqrt((x-1)(2-x))) ?
The integral of 1/(sqrt((x-1)(2-x))) is arcsin(2x-3)+C