let $f(x) = \int ^x_0 g(t) dt = \int_0 ^x \frac{t-t^{1/7} -4}{9-t}dt$
be a function find if $ \exists x_0 \in (0,9) \quad s.t\quad f(x_0)=0$ .
I'm sure that to solve this exercise will be necessary to use the followings : improper integrals, zeroes theorem,
first of all let's study the function to know in which interval i may apply the zeroes theorem.
For the fundamental theorem of calculus we know that $f'(x) = g(x) =\frac{x-x^{1/7}-4 }{9-x}$
and we need to find the intervals in which the function is monotonically increasing function in the interval $(0,9) $.
therefore $f'(x) > 0 \iff x-x^{1/7}-4 >0$
let's define $ h(x)=x-x^{1/7}-4$
we know that $ h(3) \approx 0.028 >0 , h(2.8) \approx -0.006 <0$
therefore we know that $ \exists x_0 \in (2.8 , 3): h(x_0) = 0$ because the function is continous and the values at the endpoints have opposite signs.
$ \implies f$ is surely increasing in the interval (3,9)
now we should find $ f(3) , f(9)$ to apply again the zeroes theorem. but i can't proceed. i also tried to "approximate" f(3) like that:
$ \int_0 ^x \frac{t-t^{1/7} -4}{9-t}dt$ = $ \int_0^x -1dt + \int_0 ^x \frac{5}{9-t}dt$ $ -\int_0 ^x \frac{ t^{1/7 }}{9-t}dt $
and later i tried to find two functions $\alpha(x), \beta(x) : $ $\int^x_0 \alpha(t)dt<\int_0 ^x \frac{ t^{1/7 } }{9-t}dt< \beta(x) dt $
but i managed just to find the following functions that can help me with the approximation $\frac{1}{9-t} , \frac{t}{9-t}$, but considering all the improper integrals that will appear using this "approximations" they don't work properly, because we will have different integrals that diverge with different signs, there for i am not able to approximate $f(3)$ even more i don't know how to solve and find the sign of $ f(9)$ because i'm not able to find a solution to the improper integral ( i tried using theorems for the improper integrals like the " comparison theorem" but i cannot find a solution).
Notice that if $t\leq 1$, $g(t)=\dfrac{t-t^{1/7}-4}{9-t}\leq \dfrac{2-4}{8}=-\dfrac{1}{4}$, so $$f(1)=\displaystyle\int_{0}^1g(t)dt<0$$
We also have $\displaystyle\lim_{t\to 9} \dfrac{g(t)}{h(t)}=5-9^{1/7}<+\infty$, where $h(t)=\dfrac{1}{9-t}$ (with both functions being non-negative near $9$).
Thus, $\displaystyle\int_0^9 g(t)dt$ has the same character as $\displaystyle\int_0^9 \dfrac{1}{9-t}$, that is, it diverges to $+\infty$.
In other words, $$\displaystyle\lim_{x\to 9} f(x)=\displaystyle\lim_{x\to 9} \displaystyle\int_0^x g(t)dt=+\infty $$
Then, there exists $x_0\in (0,9),$ that we may choose bigger than $1$, such that $f(x_0)>0$.
As $g$ is continuous, $f$ is too and Bolzano's Theorem assures there exists $x\in (1,x_0)$ such that $f(x)=0$.