已知(x)=dfrac (xcos x)(1+{x)^4}+(x)^2-(int )_(-1)^1f(x)dx,则(x)=dfrac (xcos x)(1+{x)^4}+(x)^2-(int )_(-1)^1f(x)dx= _.

设(x)=dfrac (1)(1+{x)^2}+sqrt (1-{x)^2}(int )_(0)^1f(x)dx, 则 (int )_(0)^1f(x)dx=设

int dfrac ({x)^3}({(1+{x)^2)}^2}dx=A.int dfrac ({x)^3}({(1+{x)^2)}^2}dx=B.int df

(int )_(-1)^1(dfrac (sin x+1)(1+{x)^2}+(x)^4)dxA (int )_(-1)^1(dfrac (sin x+1)(

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}((cos )^2x+dfrac (xcos x)(1+{cos )^2x})dx

( 单选题 ) int dfrac ({e)^x}(1+{e)^2x}dx=_________int dfrac ({e)^x}(1+{e)^2x}dx=int

int dfrac ({(1-x))^2}(x(1+{x)^2)}dx

int dfrac (x+{x)^3}(1+{x)^4}dx

(4) (int )_(0)^+infty dfrac (dx)((1+x)(1+{x)^2)}

设 gt 0, 则 =(int )_(-a)^asqrt ({a)^2-(x)^2}ln dfrac (x+sqrt {1+{x)^2}}(3)dx= __

int dfrac (1+2{x)^2}({x)^2(1+(x)^2})dx_____._____.