What is the integral of (6x+8)/(3x^2+8x-2) ?

The integral of (6x+8)/(3x^2+8x-2) is ln|9x^2+24x-6|+C

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integral of (6x+8)/(3x^2+8x-2)

\int \frac{6 x + 8}{3 x ^{2} + 8 x - 2} d x

ln 9 x^{2} + 24 x - 6 + C

Solution steps

\int \frac{6 x + 8}{3 x ^{2} + 8 x - 2} d x

Expand

\frac{6 x + 8}{3 x ^{2} + 8 x - 2} : \frac{6 x}{3 x ^{2} + 8 x - 2} + \frac{8}{3 x ^{2} + 8 x - 2}

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

\int \frac{6 x}{3 x ^{2} + 8 x - 2} d x = ln 9 x^{2} + 24 x - 6 + \frac{2 2}{11} (ln \frac{3 x + 4}{22} + 1 - ln \frac{3 x + 4}{22} - 1)

\int \frac{8}{3 x ^{2} + 8 x - 2} d x = - \frac{2 2}{11} (ln \frac{3 x + 4}{22} + 1 - ln \frac{3 x + 4}{22} - 1)

= ln 9 x^{2} + 24 x - 6 + \frac{2 2}{11} (ln \frac{3 x + 4}{22} + 1 - ln \frac{3 x + 4}{22} - 1) - \frac{2 2}{11} (ln \frac{3 x + 4}{22} + 1 - ln \frac{3 x + 4}{22} - 1)

Add similar elements:

\frac{2 2}{11} (ln \frac{3 x + 4}{22} + 1 - ln \frac{3 x + 4}{22} - 1) - \frac{2 2}{11} (ln \frac{3 x + 4}{22} + 1 - ln \frac{3 x + 4}{22} - 1) = 0

= ln 9 x^{2} + 24 x - 6

Add a constant to the solution

= ln 9 x^{2} + 24 x - 6 + C

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What is the integral of (6x+8)/(3x^2+8x-2) ?
The integral of (6x+8)/(3x^2+8x-2) is ln|9x^2+24x-6|+C