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

Let $\alpha \in (0,1)$ and $\beta = \log_{e}(1-\alpha)$. Let $P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^n}{n}$ for $x \in (0,1)$. Then the integral $\int_{0}^{\alpha} \frac{t^{50}}{1-t} dt$ is equal to

The approximate value of $\int_1^3 \frac{dx}{2+3x}$ using Simpson's rule and dividing the interval $[1,3]$ into two equal parts is

Let $[ t ]$ denote the greatest integer less than or equal to $t$. Then the value of $\int_{1}^{2} |2x - [3x]| dx$ is

The value of the integral $\int_{1}^{5}[|x-3|+|1-x|] dx$ is equal to

If $[x]$ denotes the greatest integer less than or equal to $x,$ then the value of the integral $\int_0^2 {{x^2}[x]\,dx} $ equals (in $/3$)