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

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

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

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

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

Decimal Notation

- 0.54930 \dots

Solution steps

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

Expand

\frac{- x + 1}{x ^{2} - 4} : - \frac{x}{x ^{2} - 4} + \frac{1}{x ^{2} - 4}

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

\int_{- 1}^{1} \frac{x}{x ^{2} - 4} d x = 0

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

= - 0 - \frac{1}{2} ln (3)

Simplify

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

\int_{1}^{0} (3 x^{2} + x - 5) d x

\int_{0}^{1} x (3 x^{2} - 4)^{3} d x

\int_{- π}^{π} cos (2 n x) d x

\int_{0}^{1} \frac{1}{x ^{\frac{5}{2}}} d x

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