What is the integral of ((m-15)x-(m+20))/(3x^2+x-4) ?

The integral of ((m-15)x-(m+20))/(3x^2+x-4) is

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integral of ((m-15)x-(m+20))/(3x^2+x-4)

\int \frac{( m - 15 ) x - ( m + 20 )}{3 x ^{2} + x - 4} d x

12 (m (\frac{1}{72} ln 36 x^{2} + 12 x - 48 + \frac{1}{504} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x)) - 15 (\frac{1}{72} ln 36 x^{2} + 12 x - 48 + \frac{1}{504} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x)) + \frac{m}{84} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x) + \frac{5}{21} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x)) + C

Solution steps

\int \frac{( m - 15 ) x - ( m + 20 )}{3 x ^{2} + x - 4} d x

Complete the square

3 x^{2} + x - 4 : 3 (x + \frac{1}{6})^{2} - \frac{49}{12}

= \int \frac{( m - 15 ) x - ( m + 20 )}{3 ( x + \frac{1}{6} ) ^{2} - \frac{49}{12}} d x

Apply u-substitution

: \int \frac{12 ( m ( u - \frac{1}{6} ) - 15 ( u - \frac{1}{6} ) - m - 20 )}{36 u ^{2} - 49} d u

= \int \frac{12 ( m ( u - \frac{1}{6} ) - 15 ( u - \frac{1}{6} ) - m - 20 )}{36 u ^{2} - 49} d u

Take the constant out:

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

= 12 \cdot \int \frac{m ( u - \frac{1}{6} ) - 15 ( u - \frac{1}{6} ) - m - 20}{36 u ^{2} - 49} d u

Expand

\frac{m ( u - \frac{1}{6} ) - 15 ( u - \frac{1}{6} ) - m - 20}{36 u ^{2} - 49} : \frac{m ( u - \frac{1}{6} )}{36 u ^{2} - 49} - \frac{15 ( u - \frac{1}{6} )}{36 u ^{2} - 49} - \frac{m}{36 u ^{2} - 49} - \frac{20}{36 u ^{2} - 49}

Apply the Sum Rule:

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

= 12 (\int \frac{m ( u - \frac{1}{6} )}{36 u ^{2} - 49} d u - \int \frac{15 ( u - \frac{1}{6} )}{36 u ^{2} - 49} d u - \int \frac{m}{36 u ^{2} - 49} d u - \int \frac{20}{36 u ^{2} - 49} d u)

\int \frac{m ( u - \frac{1}{6} )}{36 u ^{2} - 49} d u = m (\frac{1}{72} ln 36 u^{2} - 49 + \frac{1}{504} (ln \frac{6}{7} u + 1 - ln \frac{6}{7} u - 1))

\int \frac{15 ( u - \frac{1}{6} )}{36 u ^{2} - 49} d u = 15 (\frac{1}{72} ln 36 u^{2} - 49 + \frac{1}{504} (ln \frac{6}{7} u + 1 - ln \frac{6}{7} u - 1))

\int \frac{m}{36 u ^{2} - 49} d u = - \frac{m}{84} (ln \frac{6}{7} u + 1 - ln \frac{6}{7} u - 1)

\int \frac{20}{36 u ^{2} - 49} d u = - \frac{5}{21} (ln \frac{6}{7} u + 1 - ln \frac{6}{7} u - 1)

= 12 (m (\frac{1}{72} ln 36 u^{2} - 49 + \frac{1}{504} (ln \frac{6}{7} u + 1 - ln \frac{6}{7} u - 1)) - 15 (\frac{1}{72} ln 36 u^{2} - 49 + \frac{1}{504} (ln \frac{6}{7} u + 1 - ln \frac{6}{7} u - 1)) - (- \frac{m}{84} (ln \frac{6}{7} u + 1 - ln \frac{6}{7} u - 1)) - (- \frac{5}{21} (ln \frac{6}{7} u + 1 - ln \frac{6}{7} u - 1)))

Substitute back

u = x + \frac{1}{6}

= 12 (m (\frac{1}{72} ln 36 (x + \frac{1}{6})^{2} - 49 + \frac{1}{504} (ln \frac{6}{7} (x + \frac{1}{6}) + 1 - ln \frac{6}{7} (x + \frac{1}{6}) - 1)) - 15 (\frac{1}{72} ln 36 (x + \frac{1}{6})^{2} - 49 + \frac{1}{504} (ln \frac{6}{7} (x + \frac{1}{6}) + 1 - ln \frac{6}{7} (x + \frac{1}{6}) - 1)) - (- \frac{m}{84} (ln \frac{6}{7} (x + \frac{1}{6}) + 1 - ln \frac{6}{7} (x + \frac{1}{6}) - 1)) - (- \frac{5}{21} (ln \frac{6}{7} (x + \frac{1}{6}) + 1 - ln \frac{6}{7} (x + \frac{1}{6}) - 1)))

Simplify

12 (m (\frac{1}{72} ln 36 (x + \frac{1}{6})^{2} - 49 + \frac{1}{504} (ln \frac{6}{7} (x + \frac{1}{6}) + 1 - ln \frac{6}{7} (x + \frac{1}{6}) - 1)) - 15 (\frac{1}{72} ln 36 (x + \frac{1}{6})^{2} - 49 + \frac{1}{504} (ln \frac{6}{7} (x + \frac{1}{6}) + 1 - ln \frac{6}{7} (x + \frac{1}{6}) - 1)) - (- \frac{m}{84} (ln \frac{6}{7} (x + \frac{1}{6}) + 1 - ln \frac{6}{7} (x + \frac{1}{6}) - 1)) - (- \frac{5}{21} (ln \frac{6}{7} (x + \frac{1}{6}) + 1 - ln \frac{6}{7} (x + \frac{1}{6}) - 1))) : 12 (m (\frac{1}{72} ln 36 x^{2} + 12 x - 48 + \frac{1}{504} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x)) - 15 (\frac{1}{72} ln 36 x^{2} + 12 x - 48 + \frac{1}{504} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x)) + \frac{m}{84} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x) + \frac{5}{21} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x))

= 12 (m (\frac{1}{72} ln 36 x^{2} + 12 x - 48 + \frac{1}{504} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x)) - 15 (\frac{1}{72} ln 36 x^{2} + 12 x - 48 + \frac{1}{504} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x)) + \frac{m}{84} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x) + \frac{5}{21} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x))

Add a constant to the solution

= 12 (m (\frac{1}{72} ln 36 x^{2} + 12 x - 48 + \frac{1}{504} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x)) - 15 (\frac{1}{72} ln 36 x^{2} + 12 x - 48 + \frac{1}{504} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x)) + \frac{m}{84} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x) + \frac{5}{21} (ln \frac{8}{7} + \frac{6}{7} x - ln - \frac{6}{7} + \frac{6}{7} x)) + C

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What is the integral of ((m-15)x-(m+20))/(3x^2+x-4) ?
The integral of ((m-15)x-(m+20))/(3x^2+x-4) is