What is area y=(16)/(x^2-6x-55),y=0,x=-3,x=3 ?

The answer to area y=(16)/(x^2-6x-55),y=0,x=-3,x=3 is ln(7)

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area y=(16)/(x^2-6x-55),y=0,x=-3,x=3

area

y = \frac{16}{x ^{2} - 6 x - 55}, y = 0, 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)

\frac{d}{d x} (\frac{x ^{3}}{( 1 - x ) ^{2}})

\frac{d}{d x} (x^{2} + y^{2} - z)

\int x^{r} d x

\frac{x ^{2} - 6 x + 12}{x - 4}

What is area y=(16)/(x^2-6x-55),y=0,x=-3,x=3 ?
The answer to area y=(16)/(x^2-6x-55),y=0,x=-3,x=3 is ln(7)