What is the integral from-5 to-1 of 1/(x^2-8x+7) ?

The integral from-5 to-1 of 1/(x^2-8x+7) is (ln(3)+ln(2/3))/6

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integral from-5 to-1 of 1/(x^2-8x+7)

\int_{- 5}^{- 1} \frac{1}{x ^{2} - 8 x + 7} d x

\frac{ln ( 3 ) + ln ( \frac{2}{3} )}{6}

Decimal Notation

0.11552 \dots

Solution steps

\int_{- 5}^{- 1} \frac{1}{x ^{2} - 8 x + 7} d x

Take the partial fraction of

\frac{1}{x ^{2} - 8 x + 7} : - \frac{1}{6 ( x - 1 )} + \frac{1}{6 ( x - 7 )}

= \int_{- 5}^{- 1} - \frac{1}{6 ( x - 1 )} + \frac{1}{6 ( x - 7 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 5}^{- 1} \frac{1}{6 ( x - 1 )} d x + \int_{- 5}^{- 1} \frac{1}{6 ( x - 7 )} d x

\int_{- 5}^{- 1} \frac{1}{6 ( x - 1 )} d x = - \frac{1}{6} ln (3)

\int_{- 5}^{- 1} \frac{1}{6 ( x - 7 )} d x = \frac{1}{6} ln (\frac{2}{3})

= - (- \frac{1}{6} ln (3)) + \frac{1}{6} ln (\frac{2}{3})

Simplify

- (- \frac{1}{6} ln (3)) + \frac{1}{6} ln (\frac{2}{3}) : \frac{ln ( 3 ) + ln ( \frac{2}{3} )}{6}

= \frac{ln ( 3 ) + ln ( \frac{2}{3} )}{6}

\int_{0}^{4} (- 2 x + 2) d x

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\int_{e^{- t}}^{e^{1 - t}} ln (x) d x

What is the integral from-5 to-1 of 1/(x^2-8x+7) ?
The integral from-5 to-1 of 1/(x^2-8x+7) is (ln(3)+ln(2/3))/6