What is area y=(16)/(x^2-6x-55),x=-3,x=3 ?

The answer to area y=(16)/(x^2-6x-55),x=-3,x=3 is ln(7)

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area y=(16)/(x^2-6x-55),x=-3,x=3

area

y = \frac{16}{x ^{2} - 6 x - 55}, x = - 3, x = 3

ln (7)

Decimal Notation

1.94591 \dots

Solution steps

Apply the area formula:

\int_{- 3}^{3} \frac{16}{∣ x ^{2} - 6 x - 55 ∣} d x

Solve

\int_{- 3}^{3} \frac{16}{∣ x ^{2} - 6 x - 55 ∣} d x : ln (7)

The area is:

= ln (7)

\int \frac{x ^{2} + 132 x + 132}{x ^{3} - 4 x} d x

\int \frac{e ^{x}}{e ^{2 x} - e ^{x}} d x

\frac{\partial}{\partial z} (ln (x))

\int \frac{x}{3 x ^{2} - 1} d x

(- 10.9) (- 12.7)

What is area y=(16)/(x^2-6x-55),x=-3,x=3 ?
The answer to area y=(16)/(x^2-6x-55),x=-3,x=3 is ln(7)