Approximating the value of a definite integral I came across this question in ISI(Indian Statistical Institute) admission test
$$I=\int_2^3 \frac{dx}{\ln(x)} $$
The four options were
(A) is less than $2$ 
(B) is equal to $2$
(C) lies in the interval $(2, 3)$ 
(D) is greater than $3$
How do I go about solving this problem using high school level calculus? 
Any help is greatly appreciated! 
 A: We have
$$2\ln\left(\frac{\ln 3}{\ln 2}\right)=2\int_2^3\frac{dx}{x\ln x}\le\int_2^3\frac{dx}{\ln x}\le 3\int_2^3\frac{dx}{x\ln x}=3\ln\left(\frac{\ln 3}{\ln 2}\right)$$
Can you now choose the right option?
A: HINT: What is the smallest/largest $\dfrac1{\ln x}$ can be when $2\le x\le 3$?
A: Graphically :
Let $I=\int_2^3 \frac{1}{ln(x)}dx$
The following graph is obviously not to scale, but is useful to get yourself an idea of what we're doing.

The area of the blue rectangle* is $A_r=\frac{1}{ln(3)}$
The area of the red triangle is $A_t=\frac{1}{2} \times 1 \times (\frac{1}{ln(2)}-\frac{1}{ln(3)})$
We have $I \leqslant A_r+A_t=\frac{1}{2} \times (\frac{1}{ln(2)}+\frac{1}{ln(3)}) = \frac{ln(2)+ln(3)}{2ln(2)ln(3)} \leqslant \frac{1}{ln(2)} < 2$ since $\sqrt e <\sqrt3<1.8 < 2$
The first inequality comes from the fact that your function is convex (you can check the second derivative is positive on $[2,3]$)
edit : yeah, come to think of it, it was a bit silly,  you could just find $I\leqslant \frac{1}{ln(2)}$ immediately as other answers stated already... still $\frac{1}{2} (\frac{1}{ln(2)}+\frac{1}{ln(3)})$ is a good upper bound so I guess I'll still leave it anyway.
