Question

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. g(x) = 7x u2 − 1 u2 + 1 du 6x Hint: 7x f(u) du 6x = 0 f(u) du 6x + 7x f(u) du 0

Answer 1

It looks like you're given

$g(x)=\displaystyle\int_{6x}^{7x}\frac{u^2-1}{u^2+1}\,\mathrm du$

Then by the additivity of definite integrals this is the same as

$g(x)=\displaystyle\int_0^{7x}\frac{u^2-1}{u^2+1}\,\mathrm du-\int_0^{6x}\frac{u^2-1}{u^2+1}\,\mathrm du$

(presumably this is what the hint suggests to use)

Then by the fundamental theorem of calculus, we have

$\dfrac{\mathrm dg}{\mathrm dx}=7\dfrac{(7x)^2-1}{(7x)^2+1}-6\dfrac{(6x)^2-1}{(6x)^2+1}=\dfrac{1764x^4+169x^2-1}{1764x^4+85x^2+1}$