The value of $\int {\frac{{2dx}}{{\sqrt {1 - 4{x^2}} }}} $ is

The integral $\int {\frac{{xdx}}{{2 - {x^2} + \sqrt {2 - {x^2}} }}} $ equals

$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x=$

If $\int_1^3 x^n \sqrt{x^2-1} dx = 6$,then $n = $

If $\int \frac{x^3 \, dx}{\sqrt{1+x^2}} = a(1+x^2) \sqrt{1+x^2} + b \sqrt{1+x^2} + c$ (where $c$ is a constant of integration),then the value of $3ab$ is

The value of the integral $\int \frac{\sin(\ln(2 + 2x))}{x + 1} dx$ is