Still working some basic Fundamental Theorem of Calculus problems and I would like to know if what I do is correct when I evaluate something like $$\frac d{dx} \int_{e^{-x}}^x \ln(t^2+1)dt$$ I think this calls for both $$\int_a^b f(x)dx = \int_c^b f(x)dx-\int_c^a f(x)dx = F(b)-F(a)$$ and $$\frac d{dx} \int_a^x f(t)dt=f(x)$$ which gives me $$\frac d{dx} \int_{e^{-x}}^x \ln(t^2+1)dt = \frac d{dx} \left[\int_{e^{-x}}^c \ln(t^2+1)dt + \int_c^x \ln(t^2+1)dt\right]$$ $$= \frac d{dx} \left[\int_c^x \ln(t^2+1)dt - \int_c^{e^{-x}} \ln(t^2+1)dt\right]= \ln \left(\frac {x^2+1}{e^{-2x}+1}\right)$$ The way I see it from $\int_a^b f(x)dx = F(b)-F(a)$ could I just do it directly like this? $$\frac d{dx} \int_{e^{-x}}^x \ln(t^2+1)dt = \frac d{dx} \left[F(x)-F(e^{-x})\right]$$ $$= f(x)-f(e^{-x}) = \ln \left(\frac {x^2+1}{e^{-2x}+1}\right)$$ In general terms can I always do this $$\frac d{dx} \int_{x_1}^{x_2} f(t)dt = \frac d{dx}\left[F(x_2)-F(x_1)\right] = f(x_2)-f(x_1)$$ or would it be abusing the definition?

Thank you for taking the time to reply, it helps me a lot and I appreciate it very much!

Edit (corrections)

As pointed out in Simply Beautiful Art's comment, I should apply the chain rule, so $$\frac d{dx} \int_{e^{-x}}^x \ln(t^2+1)dt = F'(x)-F'(e^{x})\frac d{dx}e^{-x} = \ln(x^2+1) + \frac{\ln(e^{-2x}+1)}{e^x}$$

and from Ethan Bolker's answer, the last line should be changed to $$\frac d{dx} \int_{g(x)}^{h(x)} f(t)dt = \frac d{dx}\left[F(h(x)) - F(g(x))\right] =f( h(x))h'(x) - f(g(x))g'(x)$$

  • 2
    $\begingroup$ I see only one flaw, and it happens here: $$\frac d{dx}F(e^{-x})\ne f(e^{-x})$$ You've got a function inside a function, and you know what that means... chain rule! $\endgroup$ Mar 1, 2017 at 0:43
  • 2
    $\begingroup$ Arf, faster than me, I was about to say you forgot to derivate the integration bounds... $\endgroup$
    – zwim
    Mar 1, 2017 at 0:44

1 Answer 1


If you read the last line carefully you can tell that it abuses the notation (not the definition). There's nothing in that line that says $x_1$ and $x_2$ depend on $x$. In fact the integral is just a number, and its derivative is $0$.

To calculate something like $$ \frac{d}{dx}\int_c^{x^2} f(t)dt $$ you need $$ \frac{d}{dx} F(x^2). $$ That you finish with the chain rule.


You must log in to answer this question.