Originally Posted by

**Danny** The second fundemental thm of calculus is

if $\displaystyle F(x) = \int_a^x f(t)\, dt $ then $\displaystyle F'(x) = f(x).$

For your problem, split the integral up. i.e.

$\displaystyle

\int_{\sec x}^{x^2} \sqrt{1+ 2u^2}\, du = \int_{\sec x}^{a} \sqrt{1+ 2u^2}\, du + \int_{a}^{x^2} \sqrt{1+ 2u^2}\, du

$

$\displaystyle

= - \int_a^{\sec x} \sqrt{1+ 2u^2}\, du + \int_{a}^{x^2} \sqrt{1+ 2u^2}\, du,

$

where $\displaystyle a$ is a constant. Then use the chain rule on each piece.