计算(int )_(0)^+infty dfrac (dx)({x)^2+4x+3}的值是A (int )_(0)^+infty dfrac (dx)({x)^2+4x+3}B (int )_(0)^+infty dfrac (dx)({x)^2+4x+3}C (int )_(0)^+infty dfrac (dx)({x)^2+4x+3}D (int )_(0)^+infty dfrac (dx)({x)^2+4x+3}

(int )_(-infty )^+infty (x)^2(e)^-a(x^2)dx=dfrac (1)(2)sqrt (dfrac {pi )({a)^3}}

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

[例3] (int )_(0)^+infty dfrac (x{e)^-x}({(1+{e)^-x)}^2}dx

4.下列结论正确的是 ()-|||-(A) (int )_(-infty )^+infty dfrac (x)(1+{x)^2}dx=0 (B) (int )_

(int )_(0)^+infty dfrac (x{e)^-x}({(1+{e)^-x)}^2}dx= ()-|||-__

(int )_(-infty )^+infty dfrac (dx)({x)^2+2x+2}..

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

(int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=( ) .(int )_(0)^dfrac (pi {4)}dfr

[题目]计算 (int )_(1)^+infty dfrac (arctan x)({x)^2}dx

(int )_(0)^1arctan dfrac (x)(2)dx=()(int )_(0)^1arctan dfrac (x)(2)dx=()(int )_(