What is the integral from-5 to 3 of x/(sqrt(x^2-9)) ?

The integral from-5 to 3 of x/(sqrt(x^2-9)) is -4

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integral from-5 to 3 of x/(sqrt(x^2-9))

\int_{- 5}^{3} \frac{x}{x ^{2} - 9} d x

- 4

Solution steps

\int_{- 5}^{3} \frac{x}{x ^{2} - 9} d x

Find undefined (singularity) points:

x = - 3

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{- 5}^{- 3} \frac{x}{x ^{2} - 9} d x + \int_{- 3}^{3} \frac{x}{x ^{2} - 9} d x

\int_{- 5}^{- 3} \frac{x}{x ^{2} - 9} d x = - 4

\int_{- 3}^{3} \frac{x}{x ^{2} - 9} d x = 0

= - 4 + 0

Simplify

= - 4

f (x) = \frac{1 - 2 t}{3 + t}

\frac{d}{d x} (\frac{1}{10} + x e^{- x})

\frac{d}{d x} (e^{- 3 x} y)

\frac{\partial}{\partial x} (x^{0.3} y^{0.7})

y^{^{''}} + 2 y^{^{'}} + 5 y = 10 t^{2} - e^{- 2 t}

What is the integral from-5 to 3 of x/(sqrt(x^2-9)) ?
The integral from-5 to 3 of x/(sqrt(x^2-9)) is -4