What is area y=(x^2+5)/(6x-x^2),(1,5) ?

The answer to area y=(x^2+5)/(6x-x^2),(1,5) is -4+23/3 ln(5)

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area y=(x^2+5)/(6x-x^2),(1,5)

area

y = \frac{x ^{2} + 5}{6 x - x ^{2}}, (1, 5)

- 4 + \frac{23}{3} ln (5)

Decimal

8.33902 \dots

Solution steps

Apply the area formula:

\int_{1}^{5} \frac{x ^{2} + 5}{∣ 6 x - x ^{2} ∣} d x

Solve

\int_{1}^{5} \frac{x ^{2} + 5}{∣ 6 x - x ^{2} ∣} d x : - 4 + \frac{23}{3} ln (5)

The area is:

= - 4 + \frac{23}{3} ln (5)

\int - 7 lo g_{2} (x) d x

\int e^{x} (x + 1)^{2} d x

y = x^{2} - 4 x, y = 0

23 x^{3} + x^{2} - 12 x + 17

\frac{d y}{d x} = \frac{x - 3}{2 - y}

What is area y=(x^2+5)/(6x-x^2),(1,5) ?
The answer to area y=(x^2+5)/(6x-x^2),(1,5) is -4+23/3 ln(5)