What is the integral from-1 to 1 of 1-8/(9-x^2) ?

The integral from-1 to 1 of 1-8/(9-x^2) is 2-(8ln(2))/3

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integral from-1 to 1 of 1-8/(9-x^2)

\int_{- 1}^{1} 1 - \frac{8}{9 - x ^{2}} d x

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

Decimal

0.15160 \dots

Solution steps

\int_{- 1}^{1} 1 - \frac{8}{9 - x ^{2}} 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_{- 1}^{1} 1 d x - \int_{- 1}^{1} \frac{8}{9 - x ^{2}} d x

\int_{- 1}^{1} 1 d x = 2

\int_{- 1}^{1} \frac{8}{9 - x ^{2}} d x = \frac{8 ln ( 2 )}{3}

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

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What is the integral from-1 to 1 of 1-8/(9-x^2) ?
The integral from-1 to 1 of 1-8/(9-x^2) is 2-(8ln(2))/3