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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/((3-x)sqrt(2-x))

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

- 2 arctan (- x + 2) + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply u-substitution

= - \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 = arctan (v)

= - 2 arctan (v)

Substitute back

= - 2 arctan (3 - x - 1)

Simplify

= - 2 arctan (- x + 2)

Add a constant to the solution

= - 2 arctan (- x + 2) + C

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