=(int )_(-1)^1(f(x)+f(-x)+x} cdot ln (sqrt (1+{x)^2}+x)dx.

int ln (sqrt (1+{x)^2}-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 (sqrt {1+{x)^2}+sqrt (1-{x)^2}}(sqrt {1-{x)^4}}dx.

求不定积分int (x)^2(ln x-dfrac (1)(1+{x)^2})dx.求不定积分.

设 f ( x ) 在 [ -1 , 1 ] 上连续，计算=[1,|f(x)+f(-2)+z]·ln(√1+ +x)dx设f(x)在[-1,1]上连续，计算

求不定积分int dfrac (1)(2x)sqrt (ln x)dx=（）.int dfrac (1)(2x)sqrt (ln x)dx=int dfrac

第西意 不定积分-|||-(20)f(f(m(m+x)/√(1+x))dx;-|||-(10) sqrt (tan sqrt {1+{x)^2}}cdot df

int tan sqrt (1+{x)^2}cdot dfrac (xdx)(sqrt {1+{x)^2}}

(int )_(1)^edfrac (ln x)(x)dx..

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