下列式子或叙述不正确的是A.(int )_(0)^dfrac (pi {4)}(tan )^2xdx=1-dfrac (pi )(4) __-|||-__B.设函数 (int )_(0)^dfrac (pi {4)}(tan )^2xdx=1-dfrac (pi )(4) __-|||-__ 在 x = 0 点连续则 a = 1 3 C.(int )_(0)^dfrac (pi {4)}(tan )^2xdx=1-dfrac (pi )(4) __-|||-__D. (int )_(0)^dfrac (pi {4)}(tan )^2xdx=1-dfrac (pi )(4) __-|||-__

步骤 1：计算 ${\int }_{0}^{\dfrac {\pi }{4}}{\tan }^{2}xdx$
${\int }_{0}^{\dfrac {\pi }{4}}{\tan }^{2}xdx={\int }_{0}^{\dfrac {\pi }{4}}\dfrac {1-{\cos }^{2}x}{{\cos }^{2}x}dx={\int }_{0}^{\dfrac {\pi }{4}}{\sec }^{2}x-1dx={(\tan x-x)}^{\dfrac {\pi }{4}}=1-\dfrac {\pi }{4}$
步骤 2：分析函数 f(x) 在 x = 0 点的连续性
设函数 f(x)= $\left \{ \begin{matrix} \dfrac {1}{{x}^{3}}{\int }_{0}^{x}\arctan t,x\neq 0\\ a,x=0\end{matrix} \right.$ 在 x = 0 点连续，则需要满足 $\lim_{x \to 0} f(x) = a$，即 $\lim_{x \to 0} \dfrac {1}{{x}^{3}}{\int }_{0}^{x}\arctan t = a$。根据洛必达法则，可以求得 a = 1/3。
步骤 3：计算 ${\int }_{0}^{1}\dfrac {{x}^{2}}{1+{x}^{2}}dx$
${\int }_{0}^{1}\dfrac {{x}^{2}}{1+{x}^{2}}dx={\int }_{0}^{1}\dfrac {{x}^{2}+1-1}{1+{x}^{2}}dx={\int }_{0}^{1}1-\dfrac {1}{1+{x}^{2}}dx={e}^{x}-\arctan x|_{0}^{1}=1-\dfrac {\pi }{4}$
步骤 4：计算 ${\int }_{1}^{2}({x}^{2}+\dfrac {1}{{x}^{4}})dx$
${\int }_{1}^{2}({x}^{2}+\dfrac {1}{{x}^{4}})dx=\dfrac {1}{3}x^3-\dfrac {1}{3x^3}|_{1}^{2}=\dfrac {21}{8}$