∫xdx/(1+x^2) =(1/2)∫d(1+x^2)/(1+x^2) =(1/2)ln(1+x^2)+C
3个回答
这个是凑积分呀
xdx=1/2d(1+x^2)
相关问题
∫x/√1+x^2dx=____ A.1/2√1+x^2+C B.√1+x^2+C C.1/2ln(1+x^2)+C D
∫ln(1+x^2)*xdx
∫x+1/x^2-2x-3 dx= 答案为1/2【ln(x^2-2X-3)+ln(x-3/x+1)】+c
几道微积分题目(1)∫X^2/(根号1-X^2)dx(2)∫ln(1+X)dx(3)∫x*cos平方Xdx
化简求值:1/2ln(2x+2√(x^2-1))+ln(√(x+1)-√(x-1)) 2x
若∫f(x)e^1/xdx=e^1/x+C,则f(x)= A,-1/x B,-1/x^2 C,1/x D,1/x^2 请
求不定积分:∫[(1-x^2)/(1+x^2)]^(1/2)xdx
1、lim ln(1+x+2x^2)+ln(1-x+x^2)/secx-cosx
∫xsin2xdx=1/2∫x(-cos2x)′dx=1/2(-xcos2x+∫(x)′cos2xdx)=-x/2cos
ln(1+(sin(2x))^2)/ln(1+x^2) x->0