What is the integral from-1 to 1 of (x^2-1)/(x^2-4) ?

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

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integral from-1 to 1 of (x^2-1)/(x^2-4)

\int_{- 1}^{1} \frac{x ^{2} - 1}{x ^{2} - 4} d x

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

Decimal

0.35208 \dots

Solution steps

\int_{- 1}^{1} \frac{x ^{2} - 1}{x ^{2} - 4} d x

\frac{x ^{2} - 1}{x ^{2} - 4} = \frac{3}{x ^{2} - 4} + 1

= \int_{- 1}^{1} \frac{3}{x ^{2} - 4} + 1 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} \frac{3}{x ^{2} - 4} d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} \frac{3}{x ^{2} - 4} d x = - \frac{3}{2} ln (3)

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

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

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\int_{0}^{2 π} - 54 sin^{2} (t) d t

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