What is the integral from 3 to 6 of (12x)/(x^2+2x-8) ?

The integral from 3 to 6 of (12x)/(x^2+2x-8) is 2(3ln(40)+ln(5/2)-4ln(7))

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integral from 3 to 6 of (12x)/(x^2+2x-8)

\int_{3}^{6} \frac{12 x}{x ^{2} + 2 x - 8} d x

2 (3 ln (40) + ln (\frac{5}{2}) - 4 ln (7))

Solution steps

\int_{3}^{6} \frac{12 x}{x ^{2} + 2 x - 8} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \int_{3}^{6} \frac{x}{x ^{2} + 2 x - 8} d x

Complete the square

x^{2} + 2 x - 8 : (x + 1)^{2} - 9

= 12 \cdot \int_{3}^{6} \frac{x}{( x + 1 ) ^{2} - 9} d x

Apply u-substitution

: \int_{4}^{7} \frac{u - 1}{u ^{2} - 9} d u

= 12 \cdot \int_{4}^{7} \frac{u - 1}{u ^{2} - 9} d u

Expand

\frac{u - 1}{u ^{2} - 9} : \frac{u}{u ^{2} - 9} - \frac{1}{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

= 12 (\int_{4}^{7} \frac{u}{u ^{2} - 9} d u - \int_{4}^{7} \frac{1}{u ^{2} - 9} d u)

\int_{4}^{7} \frac{u}{u ^{2} - 9} d u = \frac{1}{2} (ln (40) - ln (7))

\int_{4}^{7} \frac{1}{u ^{2} - 9} d u = - \frac{ln ( \frac{5}{2} ) - ln ( 7 )}{6}

= 12 (\frac{1}{2} (ln (40) - ln (7)) - (- \frac{ln ( \frac{5}{2} ) - ln ( 7 )}{6}))

Simplify

12 (\frac{1}{2} (ln (40) - ln (7)) - (- \frac{ln ( \frac{5}{2} ) - ln ( 7 )}{6})) : 2 (3 ln (40) + ln (\frac{5}{2}) - 4 ln (7))

= 2 (3 ln (40) + ln (\frac{5}{2}) - 4 ln (7))

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What is the integral from 3 to 6 of (12x)/(x^2+2x-8) ?
The integral from 3 to 6 of (12x)/(x^2+2x-8) is 2(3ln(40)+ln(5/2)-4ln(7))