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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.

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  • $\begingroup$ @zhw. OP asked for the derivative, then someone edited the word "derivative" out (along with the picture of the problem, and typing the integral instead), let me edit it again ... $\endgroup$ – Mirko May 22 '17 at 6:28
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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 $F'(x) = \ln(x^2+1)-\ln(e^{-2x}+1)\cdot (-e^{-x})= \ln(x^2+1)+e^{-x}\cdot \ln(e^{-2x}+1)$.

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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.

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  • $\begingroup$ So I would still need to integrate by parts to find G(x) right? and then plug in the variables? $\endgroup$ – M.Bucciacchio May 22 '17 at 6:21
  • $\begingroup$ @M.Bucciacchio you do not need to integrate at all, in this type of problem. For that matter, I see you did not state a clear question, but both this answer and mine assume you were asking about $F'(x)$, so asking about a derivative, so you need not find an integral, but you need to use the FTC to find the derivative of the integral given, without finding the integral itself. Or rather, you did ask for the derivative, then someone edited the word "derivative" out ... $\endgroup$ – Mirko May 22 '17 at 6:22
  • $\begingroup$ so the answer is simply plugging in (x) and (e^-x) in place of variable (t) in the equation for F(x) $\endgroup$ – M.Bucciacchio May 22 '17 at 6:26
  • $\begingroup$ @M.Bucciacchio it is not simply that, as you need to also use the chain rule, see details in my answer $\endgroup$ – Mirko May 22 '17 at 6:27
  • $\begingroup$ yes it seems the original question was fiddled with -_- the original question which was pulled directly from my exam was "Use the Fundamental Theorem of Calculus to calculate the derivative of" which was edited by someone else $\endgroup$ – M.Bucciacchio May 22 '17 at 6:28

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