The Fundamental Theory of Calculus, Midterm Question. Use the Fundamental Theorem of Calculus to calculate the derivative of:
$$F(x) = \int_{e^{-x}}^{x} \text{ln}\left ( t^{2}+1 \right )dt$$
I need help solving this, Both my TA and Professor are unreachable and I do not understand how to use FTC to solve this. Any help or explanation would be appreciated. 
 A: Let $G(x) = \int_0^x \ln(t^2+1)dt$, then $F(x) = G(x) - G(e^{-x})$, and by the Fundamental Theorem, $G'(x)= \ln(x^2+1)$. You should be able to calculate $F'(x)$ now.
A: The FTC tell you that if $g(x)=\int_a^x f(t) dt$ (where $a$ is any constant) then $g'(x) = f(x)$. 
If we have $h(x)=\int_a^{\varphi(x)} f(t) dt$ then $h(x)=g(\varphi(x))$ and you could use the chain rule, so $h'(x)=f(\varphi(x))\cdot \varphi'(x)$. 
If $h(x)=\int_{\psi(x)}^b f(t) dt$ (where $b$ is a constant) then 
$h(x)=-\int_b^{\psi(x)} f(t) dt$ hence $h'(x)=-f(\psi(x))\cdot \psi'(x)$. 
Finally, if $h(x)=\int_{\psi(x)}^{\varphi(x)} f(t) dt$, introduce a constant $a$ and write $h(x)=\int_{\psi(x)}^a f(t) dt + \int_a^{\varphi(x)} f(t) dt = \int_a^{\varphi(x)} f(t) dt - \int_a^{\psi(x)} f(t) dt $, hence 
$h'(x)=f(\varphi(x))\cdot \varphi'(x)-f(\psi(x))\cdot \psi'(x)$. 
In your example, $F(x) = \int_{e^{-x}}^{x} \ln(t^2+1) dt$ (here $\psi(x)=e^{-x}$ and $\varphi(x)=x$), and $F'(x) = \ln(x^2+1)-\ln(e^{-2x}+1)\cdot (-e^{-x})=
\ln(x^2+1)+e^{-x}\cdot \ln(e^{-2x}+1)$. 
