I had some homework for my differential equations class, and one of the questions completely stumped me, reproduced here:
Find $\frac{dy}{dx}$ for $y = ce^{-x} + e^{-x}\int_0^x\frac{tan(t)}{t}dt$
My next line looked like
$\frac{dy}{dx} = -ce^{-x} + \frac{d}{dx}(e^{-x}\int_0^x\frac{tan(t)}{t}dt)$
and by using the Fundamental Theorem of Calculus and the Product Rule (I don't know how correctly), my next line looked like
$\frac{dy}{dx} = -ce^{-x} + (e^{-x}\frac{d}{dx}\int_0^x\frac{tan(t)}{t}dt - e^{-x}\int_0^x\frac{tan(t)}{t}dt )$
and subsequently
$\frac{dy}{dx} = -ce^{-x} + e^{-x}\frac{tan(x)}{x} - e^{-x}\int_0^x\frac{tan(t)}{t}dt$
I tried evaluating the integral, since a simple derivation made it clear that it couldn't be avoided. However, I was not able to do so, and when looking it up on www.symbolab.com, it turns out to have no elementary antiderivative/is non-integrable. Does anyone know how to solve the original question?
NOTE: the assignment deadline already passed and I have already been graded on my attempt at this question. This is not an attempt to pass off anyone's insights and work as my own.
Since $$y = c e^{-x} + e^{-x} \int_0^x \frac{\tan t}{t} \, dt,\tag1$$ as you correctly show $$y' = -ce^{-x} + e^{-x} \frac{\tan x}{x} - e^{-x} \int_0^x \frac{\tan t}{t} \, dt.\tag2$$
Now if we rearrange (1) we have $$e^{-x} \int_0^x \frac{\tan t}{t} \, dt = y - c e^{-x},$$ so substituting this result into (2) we have $$y' = -c e^{-x} + e^{-x} \frac{\tan x}{x} - (y - ce^{-x}),$$ or $$y' + y = \frac{e^{-x} \tan x}{x},$$ a first-order differential equation that is free from any integral sign and has (1) as a solution.