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

The integral of 1/(2+2x-x^2) is 1/(2sqrt(3))(ln|(x-1)/(sqrt(3))+1|-ln|(x-1)/(sqrt(3))-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/(2+2x-x^2)

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

\frac{1}{2 3} (ln \frac{x - 1}{3} + 1 - ln \frac{x - 1}{3} - 1) + C

Solution steps

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

Complete the square

2 + 2 x - x^{2} : - (x - 1)^{2} + 3

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

Apply u-substitution

: \int \frac{1}{- u ^{2} + 3} d u

= \int \frac{1}{- u ^{2} + 3} d u

Apply Integral Substitution

= \int \frac{1}{3 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{3} \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}

= \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{3} \frac{ln \frac{x - 1}{3} + 1}{2} - \frac{ln \frac{x - 1}{3} - 1}{2}

Simplify

\frac{1}{3} \frac{ln \frac{x - 1}{3} + 1}{2} - \frac{ln \frac{x - 1}{3} - 1}{2} : \frac{1}{2 3} (ln \frac{x - 1}{3} + 1 - ln \frac{x - 1}{3} - 1)

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

Add a constant to the solution

= \frac{1}{2 3} (ln \frac{x - 1}{3} + 1 - ln \frac{x - 1}{3} - 1) + C

\frac{1}{x + 1}

\int sin^{3} (x) cos^{4} (x) d x

x \to - 1 lim (\frac{x + 1}{∣ x + 1 ∣})

\int_{1}^{9} \frac{1}{x} d x

x \frac{d y}{d x} + (11 + x) y = e^{- x}

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