Let $f(x)$ be positive for all real $x$. If $I_1 = \int_{1-h}^{h} x f(x(1-x)) dx$ and $I_2 = \int_{1-h}^{h} f(x(1-x)) dx$,where $(2h-1) > 0$,then $\frac{I_1}{I_2}$ is

$\int_{0}^{\infty} \frac{x \ln x}{(1 + x^2)^2} \, dx$ is equal to

Let $f : R \rightarrow R$ be a continuous function satisfying $f(x) + f(x + k) = n$,for all $x \in R$,where $k > 0$ and $n$ is a positive integer. If $I_{1} = \int_{0}^{4nk} f(x) dx$ and $I_{2} = \int_{-k}^{3k} f(x) dx$,then:

If $(a, b)$ is the orthocentre of the triangle whose vertices are $(1, 2), (2, 3)$ and $(3, 1)$,and $I_1 = \int_{a}^{b} x \sin(4x - x^2) dx$,$I_2 = \int_{a}^{b} \sin(4x - x^2) dx$,then $36 \frac{I_1}{I_2}$ is equal to:

For any integer $n$, the value of the integral $\int_0^\pi e^{\cos^2 x} \cos^3(2n+1)x \, dx$ is:

$\int_{0}^{\infty} \log \left( x + \frac{1}{x} \right) \frac{dx}{1 + x^2}$ is equal to