What is the integral of ((m-24)x+(m+18))/(4x^2+x-3) ?

The integral of ((m-24)x+(m+18))/(4x^2+x-3) is

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integral of ((m-24)x+(m+18))/(4x^2+x-3)

\int \frac{( m - 24 ) x + ( m + 18 )}{4 x ^{2} + x - 3} d x

\frac{m}{8} ln 64 x^{2} + 16 x - 48 - 3 ln 64 x^{2} + 16 x - 48 - \frac{m}{8} (ln \frac{8}{7} x + \frac{8}{7} - ln \frac{8}{7} x - \frac{6}{7}) - 3 ln \frac{8}{7} x + \frac{8}{7} + 3 ln \frac{8}{7} x - \frac{6}{7} + C

Solution steps

\int \frac{( m - 24 ) x + m + 18}{4 x ^{2} + x - 3} d x

Complete the square

4 x^{2} + x - 3 : 4 (x + \frac{1}{8})^{2} - \frac{49}{16}

= \int \frac{( m - 24 ) x + m + 18}{4 ( x + \frac{1}{8} ) ^{2} - \frac{49}{16}} d x

Apply u-substitution

= \int \frac{2 ( 8 m u - 192 u + 7 m + 168 )}{64 u ^{2} - 49} d u

Take the constant out:

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

= 2 \cdot \int \frac{8 m u - 192 u + 7 m + 168}{64 u ^{2} - 49} d u

Expand

\frac{8 m u - 192 u + 7 m + 168}{64 u ^{2} - 49} : \frac{8 m u}{64 u ^{2} - 49} - \frac{192 u}{64 u ^{2} - 49} + \frac{7 m}{64 u ^{2} - 49} + \frac{168}{64 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

= 2 (\int \frac{8 m u}{64 u ^{2} - 49} d u - \int \frac{192 u}{64 u ^{2} - 49} d u + \int \frac{7 m}{64 u ^{2} - 49} d u + \int \frac{168}{64 u ^{2} - 49} d u)

\int \frac{8 m u}{64 u ^{2} - 49} d u = \frac{1}{16} m ln 64 u^{2} - 49

\int \frac{192 u}{64 u ^{2} - 49} d u = \frac{3}{2} ln 64 u^{2} - 49

\int \frac{7 m}{64 u ^{2} - 49} d u = - \frac{m}{16} (ln \frac{8}{7} u + 1 - ln \frac{8}{7} u - 1)

\int \frac{168}{64 u ^{2} - 49} d u = - 3 (\frac{1}{2} ln \frac{8}{7} u + 1 - \frac{1}{2} ln \frac{8}{7} u - 1)

= 2 (\frac{1}{16} m ln 64 u^{2} - 49 - \frac{3}{2} ln 64 u^{2} - 49 - \frac{m}{16} (ln \frac{8}{7} u + 1 - ln \frac{8}{7} u - 1) - 3 (\frac{1}{2} ln \frac{8}{7} u + 1 - \frac{1}{2} ln \frac{8}{7} u - 1))

Substitute back

u = x + \frac{1}{8}

= 2 (\frac{1}{16} m ln 64 (x + \frac{1}{8})^{2} - 49 - \frac{3}{2} ln 64 (x + \frac{1}{8})^{2} - 49 - \frac{m}{16} (ln \frac{8}{7} (x + \frac{1}{8}) + 1 - ln \frac{8}{7} (x + \frac{1}{8}) - 1) - 3 (\frac{1}{2} ln \frac{8}{7} (x + \frac{1}{8}) + 1 - \frac{1}{2} ln \frac{8}{7} (x + \frac{1}{8}) - 1))

Simplify

2 (\frac{1}{16} m ln 64 (x + \frac{1}{8})^{2} - 49 - \frac{3}{2} ln 64 (x + \frac{1}{8})^{2} - 49 - \frac{m}{16} (ln \frac{8}{7} (x + \frac{1}{8}) + 1 - ln \frac{8}{7} (x + \frac{1}{8}) - 1) - 3 (\frac{1}{2} ln \frac{8}{7} (x + \frac{1}{8}) + 1 - \frac{1}{2} ln \frac{8}{7} (x + \frac{1}{8}) - 1)) : \frac{m}{8} ln 64 x^{2} + 16 x - 48 - 3 ln 64 x^{2} + 16 x - 48 - \frac{m}{8} (ln \frac{8}{7} x + \frac{8}{7} - ln \frac{8}{7} x - \frac{6}{7}) - 3 ln \frac{8}{7} x + \frac{8}{7} + 3 ln \frac{8}{7} x - \frac{6}{7}

= \frac{m}{8} ln 64 x^{2} + 16 x - 48 - 3 ln 64 x^{2} + 16 x - 48 - \frac{m}{8} (ln \frac{8}{7} x + \frac{8}{7} - ln \frac{8}{7} x - \frac{6}{7}) - 3 ln \frac{8}{7} x + \frac{8}{7} + 3 ln \frac{8}{7} x - \frac{6}{7}

Add a constant to the solution

= \frac{m}{8} ln 64 x^{2} + 16 x - 48 - 3 ln 64 x^{2} + 16 x - 48 - \frac{m}{8} (ln \frac{8}{7} x + \frac{8}{7} - ln \frac{8}{7} x - \frac{6}{7}) - 3 ln \frac{8}{7} x + \frac{8}{7} + 3 ln \frac{8}{7} x - \frac{6}{7} + C

What is the integral of ((m-24)x+(m+18))/(4x^2+x-3) ?
The integral of ((m-24)x+(m+18))/(4x^2+x-3) is