What is area y=(x^2+4)/(3x-x^2),(1,2) ?

The answer to area y=(x^2+4)/(3x-x^2),(1,2) is -1+17/3 ln(2)

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area y=(x^2+4)/(3x-x^2),(1,2)

area

y = \frac{x ^{2} + 4}{3 x - x ^{2}}, (1, 2)

- 1 + \frac{17}{3} ln (2)

Solution steps

Apply the area formula:

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

Solve

\int_{1}^{2} \frac{x ^{2} + 4}{∣ 3 x - x ^{2} ∣} d x : - 1 + \frac{17}{3} ln (2)

The area is:

= - 1 + \frac{17}{3} ln (2)

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What is area y=(x^2+4)/(3x-x^2),(1,2) ?
The answer to area y=(x^2+4)/(3x-x^2),(1,2) is -1+17/3 ln(2)