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

The integral of 1/(x^2-x-1) is -1/(sqrt(5))(ln|(2x-1)/(sqrt(5))+1|-ln|(2x-1)/(sqrt(5))-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/(x^2-x-1)

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

- \frac{1}{5} (ln \frac{2 x - 1}{5} + 1 - ln \frac{2 x - 1}{5} - 1) + C

Solution steps

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

Complete the square

x^{2} - x - 1 : (x - \frac{1}{2})^{2} - \frac{5}{4}

= \int \frac{1}{( x - \frac{1}{2} ) ^{2} - \frac{5}{4}} d x

Apply u-substitution

: \int \frac{4}{4 u ^{2} - 5} d u

= \int \frac{4}{4 u ^{2} - 5} 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} - 5} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 5 ( 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 5} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 4 \cdot \frac{1}{2 5} \cdot \int \frac{1}{- ( - 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 5} (- \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 5} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 4 \cdot \frac{1}{2 5} - \frac{ln \frac{2}{5} ( x - \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{5} ( x - \frac{1}{2} ) - 1}{2}

Simplify

4 \cdot \frac{1}{2 5} - \frac{ln \frac{2}{5} ( x - \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{5} ( x - \frac{1}{2} ) - 1}{2} : - \frac{1}{5} (ln \frac{2 x - 1}{5} + 1 - ln \frac{2 x - 1}{5} - 1)

= - \frac{1}{5} (ln \frac{2 x - 1}{5} + 1 - ln \frac{2 x - 1}{5} - 1)

Add a constant to the solution

= - \frac{1}{5} (ln \frac{2 x - 1}{5} + 1 - ln \frac{2 x - 1}{5} - 1) + C

\int x arccos (2 x) d x

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

\int - 10 x^{4} e^{x^{5}} d x

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

\int 6 x (1 + 3 x^{2})^{3} d x

What is the integral of 1/(x^2-x-1) ?
The integral of 1/(x^2-x-1) is -1/(sqrt(5))(ln|(2x-1)/(sqrt(5))+1|-ln|(2x-1)/(sqrt(5))-1|)+C