What is the integral from 0 to 1 of [(5x-2)/(x^2-4)] ?

The integral from 0 to 1 of [(5x-2)/(x^2-4)] is -5ln(2)+3ln(3)

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integral from 0 to 1 of [(5x-2)/(x^2-4)]

\int_{0}^{1} [\frac{5 x - 2}{x ^{2} - 4}] d x

- 5 ln (2) + 3 ln (3)

Decimal

- 0.16989 \dots

Solution steps

\int_{0}^{1} \frac{5 x - 2}{x ^{2} - 4} d x

Expand

\frac{5 x - 2}{x ^{2} - 4} : \frac{5 x}{x ^{2} - 4} - \frac{2}{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_{0}^{1} \frac{5 x}{x ^{2} - 4} d x - \int_{0}^{1} \frac{2}{x ^{2} - 4} d x

\int_{0}^{1} \frac{5 x}{x ^{2} - 4} d x = \frac{5}{2} ln (3) - 5 ln (2)

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

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

Simplify

\frac{5}{2} ln (3) - 5 ln (2) - (- \frac{ln ( 3 )}{2}) : - 5 ln (2) + 3 ln (3)

= - 5 ln (2) + 3 ln (3)

\int_{- 2}^{2} \frac{x ^{2}}{2} - 3 d x

\int_{0}^{1} 1 + 2 x^{2} d x

\int_{0}^{0.5} 1 - \frac{x}{2} d x

\int_{0}^{2 π} - sin (t) + cos (t) d t

\int_{0}^{1} - x^{2} + 36 d x

What is the integral from 0 to 1 of [(5x-2)/(x^2-4)] ?
The integral from 0 to 1 of [(5x-2)/(x^2-4)] is -5ln(2)+3ln(3)