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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

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