What is the integral of (4x^3-x)/(6x^2-2x-1) ?

The integral of (4x^3-x)/(6x^2-2x-1) is

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integral of (4x^3-x)/(6x^2-2x-1)

\int \frac{4 x ^{3} - x}{6 x ^{2} - 2 x - 1} d x

\frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108} ln 36 x^{2} - 12 x - 6 - \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) + 1 + \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) - 1 + C

Solution steps

\int \frac{4 x ^{3} - x}{6 x ^{2} - 2 x - 1} d x

Long division

\frac{4 x ^{3} - x}{6 x ^{2} - 2 x - 1} : \frac{2 x}{3} + \frac{2}{9} + \frac{x + 2}{9 ( 6 x ^{2} - 2 x - 1 )}

= \int \frac{2 x}{3} + \frac{2}{9} + \frac{x + 2}{9 ( 6 x ^{2} - 2 x - 1 )} d x

Apply the Sum Rule:

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

= \int \frac{2 x}{3} d x + \int \frac{2}{9} d x + \int \frac{x + 2}{9 ( 6 x ^{2} - 2 x - 1 )} d x

\int \frac{2 x}{3} d x = \frac{x ^{2}}{3}

\int \frac{2}{9} d x = \frac{2}{9} x

\int \frac{x + 2}{9 ( 6 x ^{2} - 2 x - 1 )} d x = \frac{1}{108 7} (7 ln 36 x^{2} - 12 x - 6 - 13 (ln \frac{6 x - 1}{7} + 1 - ln \frac{6 x - 1}{7} - 1))

= \frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108 7} (7 ln 36 x^{2} - 12 x - 6 - 13 (ln \frac{6 x - 1}{7} + 1 - ln \frac{6 x - 1}{7} - 1))

Simplify

\frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108 7} (7 ln 36 x^{2} - 12 x - 6 - 13 (ln \frac{6 x - 1}{7} + 1 - ln \frac{6 x - 1}{7} - 1)) : \frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108} ln 36 x^{2} - 12 x - 6 - \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) + 1 + \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) - 1

= \frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108} ln 36 x^{2} - 12 x - 6 - \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) + 1 + \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) - 1

Add a constant to the solution

= \frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108} ln 36 x^{2} - 12 x - 6 - \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) + 1 + \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) - 1 + C

x \to 0 lim (\frac{( 3 x ^{2} - 2 x )}{x})

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

\int \frac{x ^{3}}{x ^{2} - 16} d x

x \to \infty lim (4 x^{4})

\frac{d y}{d x} = tan (x + y)

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