What is the integral of w/(6+4w-w^2) ?

The integral of w/(6+4w-w^2) is

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integral of w/(6+4w-w^2)

\int \frac{w}{6 + 4 w - w ^{2}} d w

- \frac{1}{2} ln w^{2} - 4 w - 6 + \frac{1}{10} ln \frac{1}{10} (w - 2) + 1 - \frac{1}{10} ln \frac{1}{10} (w - 2) - 1 + C

Solution steps

\int \frac{w}{6 + 4 w - w ^{2}} d w

Apply Integration By Parts

: \frac{1}{2 10} w (ln \frac{1}{10} (w - 2) + 1 - ln \frac{1}{10} (w - 2) - 1) - \int \frac{1}{2 10} (ln \frac{1}{10} (w - 2) + 1 - ln \frac{1}{10} (w - 2) - 1) d w

= \frac{1}{2 10} w (ln \frac{1}{10} (w - 2) + 1 - ln \frac{1}{10} (w - 2) - 1) - \int \frac{1}{2 10} (ln \frac{1}{10} (w - 2) + 1 - ln \frac{1}{10} (w - 2) - 1) d w

\int \frac{1}{2 10} (ln \frac{1}{10} (w - 2) + 1 - ln \frac{1}{10} (w - 2) - 1) d w = \frac{1}{2 10} w ln \frac{1}{10} (w - 2) + 1 - \frac{1}{2 10} w ln \frac{1}{10} (w - 2) - 1 + \frac{1}{2} ln w^{2} - 4 w - 6 - \frac{1}{10} ln \frac{1}{10} (w - 2) + 1 + \frac{1}{10} ln \frac{1}{10} (w - 2) - 1

= \frac{1}{2 10} w (ln \frac{1}{10} (w - 2) + 1 - ln \frac{1}{10} (w - 2) - 1) - (\frac{1}{2 10} w ln \frac{1}{10} (w - 2) + 1 - \frac{1}{2 10} w ln \frac{1}{10} (w - 2) - 1 + \frac{1}{2} ln w^{2} - 4 w - 6 - \frac{1}{10} ln \frac{1}{10} (w - 2) + 1 + \frac{1}{10} ln \frac{1}{10} (w - 2) - 1)

Simplify

\frac{1}{2 10} w (ln \frac{1}{10} (w - 2) + 1 - ln \frac{1}{10} (w - 2) - 1) - (\frac{1}{2 10} w ln \frac{1}{10} (w - 2) + 1 - \frac{1}{2 10} w ln \frac{1}{10} (w - 2) - 1 + \frac{1}{2} ln w^{2} - 4 w - 6 - \frac{1}{10} ln \frac{1}{10} (w - 2) + 1 + \frac{1}{10} ln \frac{1}{10} (w - 2) - 1) : - \frac{1}{2} ln w^{2} - 4 w - 6 + \frac{1}{10} ln \frac{1}{10} (w - 2) + 1 - \frac{1}{10} ln \frac{1}{10} (w - 2) - 1

= - \frac{1}{2} ln w^{2} - 4 w - 6 + \frac{1}{10} ln \frac{1}{10} (w - 2) + 1 - \frac{1}{10} ln \frac{1}{10} (w - 2) - 1

Add a constant to the solution

= - \frac{1}{2} ln w^{2} - 4 w - 6 + \frac{1}{10} ln \frac{1}{10} (w - 2) + 1 - \frac{1}{10} ln \frac{1}{10} (w - 2) - 1 + C

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