What is the integral of (2x-3)/(x^2+8x+7) ?

The integral of (2x-3)/(x^2+8x+7) is ln|x^2+8x+7|+11/6 (ln| 1/3 (x+4)+1|-ln| 1/3 (x+4)-1|)+C

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integral of (2x-3)/(x^2+8x+7)

\int \frac{2 x - 3}{x ^{2} + 8 x + 7} d x

ln x^{2} + 8 x + 7 + \frac{11}{6} (ln \frac{1}{3} (x + 4) + 1 - ln \frac{1}{3} (x + 4) - 1) + C

Solution steps

\int \frac{2 x - 3}{x ^{2} + 8 x + 7} d x

Complete the square

x^{2} + 8 x + 7 : (x + 4)^{2} - 9

= \int \frac{2 x - 3}{( x + 4 ) ^{2} - 9} d x

Apply u-substitution

: \int \frac{2 u - 11}{u ^{2} - 9} d u

= \int \frac{2 u - 11}{u ^{2} - 9} d u

Expand

\frac{2 u - 11}{u ^{2} - 9} : \frac{2 u}{u ^{2} - 9} - \frac{11}{u ^{2} - 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{2 u}{u ^{2} - 9} d u - \int \frac{11}{u ^{2} - 9} d u

\int \frac{2 u}{u ^{2} - 9} d u = ln u^{2} - 9

\int \frac{11}{u ^{2} - 9} d u = - \frac{11}{6} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1)

= ln u^{2} - 9 - (- \frac{11}{6} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1))

Substitute back

u = x + 4

= ln (x + 4)^{2} - 9 - (- \frac{11}{6} (ln \frac{x + 4}{3} + 1 - ln \frac{x + 4}{3} - 1))

Simplify

ln (x + 4)^{2} - 9 - (- \frac{11}{6} (ln \frac{x + 4}{3} + 1 - ln \frac{x + 4}{3} - 1)) : ln x^{2} + 8 x + 7 + \frac{11}{6} (ln \frac{1}{3} (x + 4) + 1 - ln \frac{1}{3} (x + 4) - 1)

= ln x^{2} + 8 x + 7 + \frac{11}{6} (ln \frac{1}{3} (x + 4) + 1 - ln \frac{1}{3} (x + 4) - 1)

Add a constant to the solution

= ln x^{2} + 8 x + 7 + \frac{11}{6} (ln \frac{1}{3} (x + 4) + 1 - ln \frac{1}{3} (x + 4) - 1) + C

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What is the integral of (2x-3)/(x^2+8x+7) ?
The integral of (2x-3)/(x^2+8x+7) is ln|x^2+8x+7|+11/6 (ln| 1/3 (x+4)+1|-ln| 1/3 (x+4)-1|)+C