What is the integral of (2x+7)/(x^2+7x+2) ?

The integral of (2x+7)/(x^2+7x+2) is

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integral of (2x+7)/(x^2+7x+2)

\int \frac{2 x + 7}{x ^{2} + 7 x + 2} d x

- \frac{1}{41} (2 x + 7) (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) + \frac{2}{41} x ln \frac{1}{41} (2 x + 7) + 1 - \frac{2}{41} x ln \frac{1}{41} (2 x + 7) - 1 + \frac{7}{41} ln \frac{1}{41} (2 x + 7) + 1 - \frac{7}{41} ln \frac{1}{41} (2 x + 7) - 1 + ln (2 x + 7 + 41 2 x + 7 - 41) + C

Solution steps

\int \frac{2 x + 7}{x ^{2} + 7 x + 2} d x

Apply Integration By Parts

: - \frac{1}{41} (2 x + 7) (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) - \int - \frac{2}{41} (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) d x

= - \frac{1}{41} (2 x + 7) (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) - \int - \frac{2}{41} (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) d x

\int - \frac{2}{41} (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) d x = - \frac{2}{41} x ln \frac{1}{41} (2 x + 7) + 1 + \frac{2}{41} x ln \frac{1}{41} (2 x + 7) - 1 - \frac{7}{41} ln \frac{1}{41} (2 x + 7) + 1 + \frac{7}{41} ln \frac{1}{41} (2 x + 7) - 1 - ln (2 x + 7 + 41 2 x + 7 - 41)

= - \frac{1}{41} (2 x + 7) (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) - (- \frac{2}{41} x ln \frac{1}{41} (2 x + 7) + 1 + \frac{2}{41} x ln \frac{1}{41} (2 x + 7) - 1 - \frac{7}{41} ln \frac{1}{41} (2 x + 7) + 1 + \frac{7}{41} ln \frac{1}{41} (2 x + 7) - 1 - ln (2 x + 7 + 41 2 x + 7 - 41))

Simplify

- \frac{1}{41} (2 x + 7) (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) - (- \frac{2}{41} x ln \frac{1}{41} (2 x + 7) + 1 + \frac{2}{41} x ln \frac{1}{41} (2 x + 7) - 1 - \frac{7}{41} ln \frac{1}{41} (2 x + 7) + 1 + \frac{7}{41} ln \frac{1}{41} (2 x + 7) - 1 - ln (2 x + 7 + 41 2 x + 7 - 41)) : - \frac{1}{41} (2 x + 7) (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) + \frac{2}{41} x ln \frac{1}{41} (2 x + 7) + 1 - \frac{2}{41} x ln \frac{1}{41} (2 x + 7) - 1 + \frac{7}{41} ln \frac{1}{41} (2 x + 7) + 1 - \frac{7}{41} ln \frac{1}{41} (2 x + 7) - 1 + ln (2 x + 7 + 41 2 x + 7 - 41)

= - \frac{1}{41} (2 x + 7) (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) + \frac{2}{41} x ln \frac{1}{41} (2 x + 7) + 1 - \frac{2}{41} x ln \frac{1}{41} (2 x + 7) - 1 + \frac{7}{41} ln \frac{1}{41} (2 x + 7) + 1 - \frac{7}{41} ln \frac{1}{41} (2 x + 7) - 1 + ln (2 x + 7 + 41 2 x + 7 - 41)

Add a constant to the solution

= - \frac{1}{41} (2 x + 7) (ln \frac{1}{41} (2 x + 7) + 1 - ln \frac{1}{41} (2 x + 7) - 1) + \frac{2}{41} x ln \frac{1}{41} (2 x + 7) + 1 - \frac{2}{41} x ln \frac{1}{41} (2 x + 7) - 1 + \frac{7}{41} ln \frac{1}{41} (2 x + 7) + 1 - \frac{7}{41} ln \frac{1}{41} (2 x + 7) - 1 + ln (2 x + 7 + 41 2 x + 7 - 41) + C

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What is the integral of (2x+7)/(x^2+7x+2) ?
The integral of (2x+7)/(x^2+7x+2) is