What is the integral of (8-3x)/(10x^2+13x-3) ?

The integral of (8-3x)/(10x^2+13x-3) is

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integral of (8-3x)/(10x^2+13x-3)

\int \frac{8 - 3 x}{10 x ^{2} + 13 x - 3} d x

- \frac{8}{17} (ln \frac{20}{17} x + \frac{30}{17} - ln \frac{20}{17} x - \frac{4}{17}) - 6 (\frac{1}{40} ln 400 x^{2} + 520 x - 120 + \frac{13}{680} (ln \frac{30}{17} + \frac{20}{17} x - ln - \frac{4}{17} + \frac{20}{17} x)) + C

Solution steps

\int \frac{8 - 3 x}{10 x ^{2} + 13 x - 3} d x

Expand

\frac{8 - 3 x}{10 x ^{2} + 13 x - 3} : \frac{8}{10 x ^{2} + 13 x - 3} - \frac{3 x}{10 x ^{2} + 13 x - 3}

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{8}{10 x ^{2} + 13 x - 3} d x - \int \frac{3 x}{10 x ^{2} + 13 x - 3} d x

\int \frac{8}{10 x ^{2} + 13 x - 3} d x = - \frac{8}{17} (ln \frac{20}{17} x + \frac{30}{17} - ln \frac{20}{17} x - \frac{4}{17})

\int \frac{3 x}{10 x ^{2} + 13 x - 3} d x = 6 (\frac{1}{40} ln 400 x^{2} + 520 x - 120 + \frac{13}{680} (ln \frac{30}{17} + \frac{20}{17} x - ln - \frac{4}{17} + \frac{20}{17} x))

= - \frac{8}{17} (ln \frac{20}{17} x + \frac{30}{17} - ln \frac{20}{17} x - \frac{4}{17}) - 6 (\frac{1}{40} ln 400 x^{2} + 520 x - 120 + \frac{13}{680} (ln \frac{30}{17} + \frac{20}{17} x - ln - \frac{4}{17} + \frac{20}{17} x))

Add a constant to the solution

= - \frac{8}{17} (ln \frac{20}{17} x + \frac{30}{17} - ln \frac{20}{17} x - \frac{4}{17}) - 6 (\frac{1}{40} ln 400 x^{2} + 520 x - 120 + \frac{13}{680} (ln \frac{30}{17} + \frac{20}{17} x - ln - \frac{4}{17} + \frac{20}{17} x)) + C

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What is the integral of (8-3x)/(10x^2+13x-3) ?
The integral of (8-3x)/(10x^2+13x-3) is