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

The integral of 1/(3x-x^2-2) is ln|2x-2|-ln|2x-4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{1}{3 x - x ^{2} - 2} d x

ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣ + C

Solution steps

\int \frac{1}{3 x - x ^{2} - 2} d x

Complete the square

3 x - x^{2} - 2 : - (x - \frac{3}{2})^{2} + \frac{1}{4}

= \int \frac{1}{- ( x - \frac{3}{2} ) ^{2} + \frac{1}{4}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

= 4 \cdot \frac{1}{2} \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}

= 4 \cdot \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= 4 \cdot \frac{1}{2} (\frac{ln 2 ( x - \frac{3}{2} ) + 1}{2} - \frac{ln 2 ( x - \frac{3}{2} ) - 1}{2})

Simplify

4 \cdot \frac{1}{2} (\frac{ln 2 ( x - \frac{3}{2} ) + 1}{2} - \frac{ln 2 ( x - \frac{3}{2} ) - 1}{2}) : ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣

= ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣

Add a constant to the solution

= ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣ + C

\int 3 cot^{5} (θ) sin^{4} (θ) d θ

x^{2} - 2 x + 1

arctan (x - \frac{π}{2})

\int x^{3} x^{2} + 35 d x

What is the integral of 1/(3x-x^2-2) ?
The integral of 1/(3x-x^2-2) is ln|2x-2|-ln|2x-4|+C