题目
第四章 不正积分-|||-(20)arctan√x/dx;-|||-(10) sqrt (18sqrt {1+{x)^2}}cdot dfrac (xsqrt {x)}(sqrt {1+{x)^2}}-|||-(22) int dfrac (dx)(sin xcos x)-|||-(21) int dfrac (1+ln x)({(xln x))^2}dx-|||-(24)∫cos^3xdx-|||-(23) int dfrac (ln sin x dx)(cos xsin x)dx-|||-(26)∫sin2xcos33xdx;-|||-(25) int (cos )^2(omega t+varphi )dt;-|||-(28)int sin 5xsin 7xdx;-|||-(27) int cos xcos dfrac (x)(2)dx;-|||-(29) int (sin )^3xsec xdx;-|||-(30) int dfrac (dx)({e)^x+(e)^-x}-|||-(31) int dfrac (1-x)(sqrt {9-4{x)^2}}dx-|||-(32) int dfrac ({x)^3}(9+{x)^2}dx;-|||-(33) int dfrac (dx)(2{x)^2-1};-|||-(34) int dfrac (dx)((x+1)(x-2));-|||-(35) int dfrac (x)({x)^2-x-2}dx;-|||-(36) int dfrac ({x)^2dx}(sqrt {{a)^2-(x)^2}}(agt 0)-|||-(37) int dfrac (dx)(xsqrt {{x)^2-1}}-|||-(38) int dfrac (dx)(sqrt {{({x)^2+1)}^3}}-|||-(39) int dfrac (sqrt {{x)^2-9}}(x)dx-|||-(40) int dfrac (dx)(1+sqrt {2x)}-|||-(41) int dfrac (dx{d)^2}(1+sqrt {1-{x)^2}}-|||-(42) int dfrac (dx)(x+sqrt {1-{x)^2}}-|||-(43) int dfrac (x-1)({x)^2+2x+3}dx;-|||-(44) int dfrac ({x)^3+1}({({x)^2+1)}^2}dx.-|||-前面我们在复合函数求导法则的基础上,得到了换元积分法.-|||-个函数乘积的求导法则,来推得另一个求积分的基本方法一-|||-设函数 u=u(x) 及 v=v(x) 具有连续导数,则两个函数乘积的-|||-(uv)'=u'v+uv'
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