Using FTOC to solve a definite integral with a function as its upper bound Let $h = e^x$ $$\frac{d}{dx} \int_2^h ln(e^{2t} + 1) dt$$ Okay, so I believe that I can assume $ln(e^{2t} + 1) = f(t)$ And then $$\int_2^h f...--prophetes.ai

Using FTOC to solve a definite integral with a function as its upper bound Let $h = e^x$ $$\frac{d}{dx} \int_2^h ln(e^{2t} + 1) dt$$ Okay, so I believe that I can assume $ln(e^{2t} + 1) = f(t)$ And then $$\int_2^h f(t) dt = F(x)$$ Now to solve this by FTOC, I'm not entirely sure but normally I would just substitute $e^x$ in to $f(t)$ to get $ln(e^{2e^x}+1)$ and of which I differentiate but that is against the entire idea of the FTOC I believe? I think my understanding of the FTOC is hazy but if $h$ were not a function and just x, I would simply substitute $x$ for $t$. How does this differ from the fact that $h$ is a function?

As N74 mentions, we should actually have

$$\int_2^h f(t) dt =F(h)-F(2)=F(e^x)-F(2)$$

Whereupon the derivative would be taken with chain rule.