Application of fundamental calculus theorem i want to calculate the derivative of:
$$g(x)= x^2 \int_{{x}}^{{1}} \sqrt{u^2+1} $$
I want to apply the fundamental calculus theorem so i do:
$$g(x)= x^2 -\int_{{1}}^{{x}} \sqrt{u^2+1} $$
Then i solve in this way:
$$(x^2) -\sqrt{x^2+1}  $$
Is this process right?

From
$$g(x)= x^2 \times\int_{{x}}^{{1}} \sqrt{u^2+1} \:du$$
one may write
$$g(x)= x^2\times\left(- \int_1^x\sqrt{u^2+1} \:du\right)$$
then by differentiating, using $$(uv)'=u'v+uv'$$ one has
$$
g'(x)= 2x\times\left(- \int_1^x\sqrt{u^2+1} \:du\right)+x^2\times \left(- \sqrt{x^2+1}\right).
$$
