What is the integral of 6/(x^2-x-2) ?

The integral of 6/(x^2-x-2) is -4(1/2 ln| 2/3+2/3 x|-1/2 ln|-4/3+2/3 x|)+C

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integral of 6/(x^2-x-2)

\int \frac{6}{x ^{2} - x - 2} d x

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

Solution steps

\int \frac{6}{x ^{2} - x - 2} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x ^{2} - x - 2} d x

Complete the square

x^{2} - x - 2 : (x - \frac{1}{2})^{2} - \frac{9}{4}

= 6 \cdot \int \frac{1}{( x - \frac{1}{2} ) ^{2} - \frac{9}{4}} d x

Apply u-substitution

: \int \frac{4}{4 u ^{2} - 9} d u

= 6 \cdot \int \frac{4}{4 u ^{2} - 9} d u

Take the constant out:

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

= 6 \cdot 4 \cdot \int \frac{1}{4 u ^{2} - 9} d u

Apply Integral Substitution

= 6 \cdot 4 \cdot \int \frac{1}{6 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 6 \cdot 4 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 6 \cdot 4 \cdot \frac{1}{6} \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

= 6 \cdot 4 \cdot \frac{1}{6} (- \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}

= 6 \cdot 4 \cdot \frac{1}{6} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 6 \cdot 4 \cdot \frac{1}{6} (- (\frac{ln \frac{2}{3} ( x - \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( x - \frac{1}{2} ) - 1}{2}))

Simplify

6 \cdot 4 \cdot \frac{1}{6} (- (\frac{ln \frac{2}{3} ( x - \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( x - \frac{1}{2} ) - 1}{2})) : - 4 (\frac{1}{2} ln \frac{2}{3} + \frac{2}{3} x - \frac{1}{2} ln - \frac{4}{3} + \frac{2}{3} x)

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

Add a constant to the solution

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

y^{^{'}} (1 - x y^{1985}) = y^{1986}

\int \frac{5 x ^{3} - 3 x}{x} d x

\int_{22}^{4} \frac{1}{t ^{3} t ^{2} - 4} d t

y^{^{'}} + y cos (x) = 0

x + 1

What is the integral of 6/(x^2-x-2) ?
The integral of 6/(x^2-x-2) is -4(1/2 ln| 2/3+2/3 x|-1/2 ln|-4/3+2/3 x|)+C