What is the integral of (3x+8)/(9x^2-3x-1) ?

The integral of (3x+8)/(9x^2-3x-1) is

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integral of (3x+8)/(9x^2-3x-1)

\int \frac{3 x + 8}{9 x ^{2} - 3 x - 1} d x

4 (3 (\frac{1}{72} ln 36 x^{2} - 12 x - 4 - \frac{1}{72 5} (ln \frac{6 x - 1}{5} + 1 - ln \frac{6 x - 1}{5} - 1)) - \frac{2}{3 5} (ln \frac{6 x - 1}{5} + 1 - ln \frac{6 x - 1}{5} - 1)) + C

Solution steps

\int \frac{3 x + 8}{9 x ^{2} - 3 x - 1} d x

Complete the square

9 x^{2} - 3 x - 1 : 9 (x - \frac{1}{6})^{2} - \frac{5}{4}

= \int \frac{3 x + 8}{9 ( x - \frac{1}{6} ) ^{2} - \frac{5}{4}} d x

Apply u-substitution

= \int \frac{4 ( 3 ( u + \frac{1}{6} ) + 8 )}{36 u ^{2} - 5} d u

Take the constant out:

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

= 4 \cdot \int \frac{3 ( u + \frac{1}{6} ) + 8}{36 u ^{2} - 5} d u

Expand

\frac{3 ( u + \frac{1}{6} ) + 8}{36 u ^{2} - 5} : \frac{3 ( u + \frac{1}{6} )}{36 u ^{2} - 5} + \frac{8}{36 u ^{2} - 5}

Apply the Sum Rule:

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

= 4 (\int \frac{3 ( u + \frac{1}{6} )}{36 u ^{2} - 5} d u + \int \frac{8}{36 u ^{2} - 5} d u)

\int \frac{3 ( u + \frac{1}{6} )}{36 u ^{2} - 5} d u = 3 (\frac{1}{72} ln 36 u^{2} - 5 - \frac{1}{72 5} (ln \frac{6}{5} u + 1 - ln \frac{6}{5} u - 1))

\int \frac{8}{36 u ^{2} - 5} d u = - \frac{2}{3 5} (ln \frac{6}{5} u + 1 - ln \frac{6}{5} u - 1)

= 4 (3 (\frac{1}{72} ln 36 u^{2} - 5 - \frac{1}{72 5} (ln \frac{6}{5} u + 1 - ln \frac{6}{5} u - 1)) - \frac{2}{3 5} (ln \frac{6}{5} u + 1 - ln \frac{6}{5} u - 1))

Substitute back

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

= 4 (3 (\frac{1}{72} ln 36 (x - \frac{1}{6})^{2} - 5 - \frac{1}{72 5} (ln \frac{6}{5} (x - \frac{1}{6}) + 1 - ln \frac{6}{5} (x - \frac{1}{6}) - 1)) - \frac{2}{3 5} (ln \frac{6}{5} (x - \frac{1}{6}) + 1 - ln \frac{6}{5} (x - \frac{1}{6}) - 1))

Simplify

4 (3 (\frac{1}{72} ln 36 (x - \frac{1}{6})^{2} - 5 - \frac{1}{72 5} (ln \frac{6}{5} (x - \frac{1}{6}) + 1 - ln \frac{6}{5} (x - \frac{1}{6}) - 1)) - \frac{2}{3 5} (ln \frac{6}{5} (x - \frac{1}{6}) + 1 - ln \frac{6}{5} (x - \frac{1}{6}) - 1)) : 4 (3 (\frac{1}{72} ln 36 x^{2} - 12 x - 4 - \frac{1}{72 5} (ln \frac{6 x - 1}{5} + 1 - ln \frac{6 x - 1}{5} - 1)) - \frac{2}{3 5} (ln \frac{6 x - 1}{5} + 1 - ln \frac{6 x - 1}{5} - 1))

= 4 (3 (\frac{1}{72} ln 36 x^{2} - 12 x - 4 - \frac{1}{72 5} (ln \frac{6 x - 1}{5} + 1 - ln \frac{6 x - 1}{5} - 1)) - \frac{2}{3 5} (ln \frac{6 x - 1}{5} + 1 - ln \frac{6 x - 1}{5} - 1))

Add a constant to the solution

= 4 (3 (\frac{1}{72} ln 36 x^{2} - 12 x - 4 - \frac{1}{72 5} (ln \frac{6 x - 1}{5} + 1 - ln \frac{6 x - 1}{5} - 1)) - \frac{2}{3 5} (ln \frac{6 x - 1}{5} + 1 - ln \frac{6 x - 1}{5} - 1)) + C

What is the integral of (3x+8)/(9x^2-3x-1) ?
The integral of (3x+8)/(9x^2-3x-1) is