Question

how to i evalute the definite inegral.
the integral of 3 to 6 x/(3 squareroot (x^2-8))dx

Answer 1

\[\int\limits_{3}^{6}x /3\sqrt{x^2-8} dx\]

Answer 2

Is the square root in the denominator?

Answer 3

yes

Answer 4

\[\large \int\limits_3^6 \frac{1}{3\sqrt{x^2-8}}x \; dx\]I moved the x off to the side just because it will make it a little easier to see our U-substitution.

If we let,\[\large u=x^2-8\]What do we get for du?
I think it will be something very similar to that x dx yes? :o

Answer 5

du would = 2x so du=2x