If $\int_{1}^{k}(3x^{2}+2x+1)dx=11$,then $k=$

$\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x=$

Suppose that $F(x)$ is an antiderivative of $f(x) = \frac{\sin x}{x}$,$x > 0$. Then $\int_{1}^{3} \frac{\sin 2x}{x} dx$ can be expressed as:

$\int_{0}^{1} \frac{d}{dx} \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$ is equal to

The value of the definite integral,$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin 5x}}{{\sin x}}\,dx} $ is

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is