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

The integral from-2 to-1 of (x-1)/(2x^2+8x) is (5ln(3)-4ln(2))/8

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

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

\frac{5 ln ( 3 ) - 4 ln ( 2 )}{8}

Solution steps

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

Expand

\frac{x - 1}{2 x ^{2} + 8 x} : \frac{x}{2 x ^{2} + 8 x} - \frac{1}{2 x ^{2} + 8 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_{- 2}^{- 1} \frac{x}{2 x ^{2} + 8 x} d x - \int_{- 2}^{- 1} \frac{1}{2 x ^{2} + 8 x} d x

\int_{- 2}^{- 1} \frac{x}{2 x ^{2} + 8 x} d x = \frac{ln ( 3 ) - ln ( 2 )}{2}

\int_{- 2}^{- 1} \frac{1}{2 x ^{2} + 8 x} d x = - \frac{1}{8} ln (3)

= \frac{ln ( 3 ) - ln ( 2 )}{2} - (- \frac{1}{8} ln (3))

Simplify

\frac{ln ( 3 ) - ln ( 2 )}{2} - (- \frac{1}{8} ln (3)) : \frac{5 ln ( 3 ) - 4 ln ( 2 )}{8}

= \frac{5 ln ( 3 ) - 4 ln ( 2 )}{8}

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What is the integral from-2 to-1 of (x-1)/(2x^2+8x) ?
The integral from-2 to-1 of (x-1)/(2x^2+8x) is (5ln(3)-4ln(2))/8