What is the integral of (28t+28)/(14t^2+28t-9) ?

The integral of (28t+28)/(14t^2+28t-9) is ln|14t^2+28t-9|+C

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integral of (28t+28)/(14t^2+28t-9)

\int \frac{28 t + 28}{14 t ^{2} + 28 t - 9} d t

ln 14 t^{2} + 28 t - 9 + C

Solution steps

\int \frac{28 t + 28}{14 t ^{2} + 28 t - 9} d t

Expand

\frac{28 t + 28}{14 t ^{2} + 28 t - 9} : \frac{28 t}{14 t ^{2} + 28 t - 9} + \frac{28}{14 t ^{2} + 28 t - 9}

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{28 t}{14 t ^{2} + 28 t - 9} d t + \int \frac{28}{14 t ^{2} + 28 t - 9} d t

\int \frac{28 t}{14 t ^{2} + 28 t - 9} d t = ln 14 t^{2} + 28 t - 9 + \frac{14}{23} ln \frac{14}{23} (t + 1) + 1 - \frac{14}{23} ln \frac{14}{23} (t + 1) - 1

\int \frac{28}{14 t ^{2} + 28 t - 9} d t = - \frac{14}{23} (ln \frac{14}{23} (t + 1) + 1 - ln \frac{14}{23} (t + 1) - 1)

= ln 14 t^{2} + 28 t - 9 + \frac{14}{23} ln \frac{14}{23} (t + 1) + 1 - \frac{14}{23} ln \frac{14}{23} (t + 1) - 1 - \frac{14}{23} (ln \frac{14}{23} (t + 1) + 1 - ln \frac{14}{23} (t + 1) - 1)

Simplify

ln 14 t^{2} + 28 t - 9 + \frac{14}{23} ln \frac{14}{23} (t + 1) + 1 - \frac{14}{23} ln \frac{14}{23} (t + 1) - 1 - \frac{14}{23} (ln \frac{14}{23} (t + 1) + 1 - ln \frac{14}{23} (t + 1) - 1) : ln 14 t^{2} + 28 t - 9

= ln 14 t^{2} + 28 t - 9

Add a constant to the solution

= ln 14 t^{2} + 28 t - 9 + C

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What is the integral of (28t+28)/(14t^2+28t-9) ?
The integral of (28t+28)/(14t^2+28t-9) is ln|14t^2+28t-9|+C