Question

Find the derivative of the function.

http://www.webassign.net/cgi-bin/symimage.cgi?expr=g%28x%29%3Dint_%283%20x%29%5E%284%20x%29%20%28u%5E2-1%29%2F%28u%5E2%2B1%29%20du

Answer 1

Let \[f(x) = \frac{x^2 -1}{x^2 + 1},\]then\[g(x) = \int_{3x}^{4x}f(u)du.\]By the FTC:\[g'(x) = f(4x)(4x)'-f(3x)(3x)' = 4\frac{16x^2 - 1}{16x^2 + 1}-3\frac{9x^2 - 1}{9x^2 + 1} = \frac{144x^4+49x^2 - 1}{144x^4 +25x^2 +1}.\]

Answer 2

THANK YOU SO MUCH :D

Answer 3

\[\int\limits_{3 x}^{4 x} \frac{u^2-1}{u^2+1} \, du = x+2 \tan ^{-1}(3 x)-2 \tan ^{-1}(4 x) \]