What is the integral of (3x)/(x^2-x-6) ?

The integral of (3x)/(x^2-x-6) is 6(1/4 ln|4x^2-4x-24|-1/20 (ln| 4/5+2/5 x|-ln|-6/5+2/5 x|))+C

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integral of (3x)/(x^2-x-6)

\int \frac{3 x}{x ^{2} - x - 6} d x

6 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x)) + C

Solution steps

\int \frac{3 x}{x ^{2} - x - 6} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{2} - x - 6} d x

Complete the square

x^{2} - x - 6 : (x - \frac{1}{2})^{2} - \frac{25}{4}

= 3 \cdot \int \frac{x}{( x - \frac{1}{2} ) ^{2} - \frac{25}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u + 1 )}{4 u ^{2} - 25} d u

= 3 \cdot \int \frac{2 ( 2 u + 1 )}{4 u ^{2} - 25} d u

Take the constant out:

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

= 3 \cdot 2 \cdot \int \frac{2 u + 1}{4 u ^{2} - 25} d u

Expand

\frac{2 u + 1}{4 u ^{2} - 25} : \frac{2 u}{4 u ^{2} - 25} + \frac{1}{4 u ^{2} - 25}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot 2 (\int \frac{2 u}{4 u ^{2} - 25} d u + \int \frac{1}{4 u ^{2} - 25} d u)

\int \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{4} ln 4 u^{2} - 25

\int \frac{1}{4 u ^{2} - 25} d u = - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

= 3 \cdot 2 (\frac{1}{4} ln 4 u^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1))

Substitute back

u = x - \frac{1}{2}

= 3 \cdot 2 (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} (x - \frac{1}{2}) + 1 - ln \frac{2}{5} (x - \frac{1}{2}) - 1))

Simplify

3 \cdot 2 (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} (x - \frac{1}{2}) + 1 - ln \frac{2}{5} (x - \frac{1}{2}) - 1)) : 6 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x))

= 6 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x))

Add a constant to the solution

= 6 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x)) + C

\int_{0}^{1} 3 x (x^{2} + 1)^{4} d x

x^{2} \frac{d y}{d x} + 3 x y = ln (x)

x y = cot (x y)

\int \frac{x}{( z ^{2} + x ^{2} ) ^{\frac{3}{2}}} d x

y = \frac{3}{x}

What is the integral of (3x)/(x^2-x-6) ?
The integral of (3x)/(x^2-x-6) is 6(1/4 ln|4x^2-4x-24|-1/20 (ln| 4/5+2/5 x|-ln|-6/5+2/5 x|))+C