Prove $x=\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \mathrm{d}t $ for some $x>1$ Prove that $$x=\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \mathrm{d}t $$ for some $x>1.$
The only idea I have is to calculate the derivative of $\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \ \mathrm{d}t $. First, I let $y=\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \mathrm{d}t -x$ , then I calculate $y'(x)$ , but $y(x)$ isn't a monotonous function, so I got stuck.
Can anyone help or give me a hint on how to approach this problem?
Thank you very much guys
 A: At x = 1 then
$$\int_{1}^{x} cos \frac{1}{t} ⋅ ln(3 - \frac{1}{t})dt = 0 \lt x $$
What does the integrand approach as t approaches infinity?
Can you see what the integral approaches in terms of x by considering the graph of this?
The integral is continuous for x $\geq$ 1. Can you see how to prove it from here?
A: Let $I(x) = \int_1^x \cos(\frac1t) \ln(3 - \frac1t) dt$. As Ben suggested, we will use the intermediate value theorem.
First, $I(1) = 0$. Thus, $I(1) - 1 < 0$.
Then, for $x > 1$ :
$$I(x) = \ln(3) \int_1^x \cos(1/t) dt + \int_1^x \cos(1/t) \ln(1 - 1/3t) dt$$
Both integrals diverge, and the integrand are positive, thus :
$$I(x) = \ln(3) \int_1^x 1 \times dt + o(x) - \int_1^x 1 \times \frac1{3t} dt + o(\ln(x)) = \ln(3) x + o(x) \quad [x \to \infty]$$
Finally :
$$I(x) - x = (\ln(3)-1) x + o(x) \geq 0.005 x + o(x) \to \infty \quad [x \to \infty]$$
Thus, there exist $x_0 > 1$ such that $I(x_0) - x_0 > 0$. Moreover $x \mapsto I(x)$ is continuous (antiderivative of a continous function). The intermediate value theorem gives the existence of $x$ such that $I(x) = x$.
A: This can only be done numerically.
This is how it looks:

The plot shows $x$ and the integral. Blue is the integral calculated numerically and the sand colored graph is $x$.
This solution is graphical methodology, plot and adopted the plot interval. This is done with Mathematica and the built-in NIntegrate with default settings.
In Mathematica a numerical solution input is
FindRoot[NIntegrate[Cos[1/t] Log[3 - 1/t], {t, 1, x}] == x, {x, 
   27.22}, AccuracyGoal -> 8, PrecisionGoal -> 8][[1, 2]]

The output is $27.2244$. This result is properly bigger than one. The difference between x and the integral is smaller and $10^{-8}$ at this point.
At the start the value of the integral is less than $x$. The increase is already faster than $x$. The value of the integral grow faster than x for positive value bigger than 1.
This uses a famous algorithm called fixpoint iteration. The rewrite into the fixpoint form and the iteration are done internally and optimized. This does not find a more exact value than the fiven one. This fixpoint iteration does involve the first derivative.
This is only a numerical prove open to verification. But this verification can only be done numerically or graphically.
Plot of the both function on a larger interval:

Plot of the crossing of both functions represented by the difference:
.
