Evaluate the definite integral $\int_{2}^{3} \frac{1}{x} d x$.

Let $I_{1}=\int_{0}^{n}[x] d x$ and $I_{2}=\int_{0}^{n}\{x\} d x,$ where $[x]$ and $\{x\}$ are the integral and fractional parts of $x$ respectively, and $n \in N-\{1\}.$ Then, $I_{1} / I_{2}$ is equal to

Let $l = \mathop {Lim}\limits_{x \to \infty } \int\limits_x^{2x} \frac{dt}{t}$ and $m = \mathop {Lim}\limits_{x \to \infty } \frac{1}{x \ln x} \int\limits_1^x \ln t \, dt$. Then the correct statement is:

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^9 x \, dx =$

Let $I = \int_a^b (x^4 - 2x^2) dx$. If $I$ is minimum,then the ordered pair $(a, b)$ is

The value of $\int_{0}^{\pi / 2} \frac{\cos 3 x+1}{2 \cos x-1} d x$ is