∫xsin(x^2+1)dx
因为d(x^2)=2xdx
所以上面式子
=(1/2)*∫sin(x^2+1)d（x^2)
=(1/2)*∫sin(x^2+1)d（x^2+1)
=(1/2)*(-cos(x^2+1))
=-[cos(x^2+1)]/2
∫xsin(x^2+1)dx
1个回答
相关问题
-
∫xsin(x+1)dx
-
1.∫x/根号下（1+x) dx 2.∫x^2e^x dx 3.∫xarctanx dx 4.∫xsin2x dx
-
∫2xsin(x^2+3)dx
-
1、∫x√x^2+1dx2、∫e^xsin(e^x-2)dx3、∫(lnx/x)dx4、∫(1/x^2)乘tan(1/x
-
不定积分xsin(2+x^2)dx和(x-1)(x^2-2x+3)^10dx怎么解?
-
1.求不定积分∫（x^2）dx/（x-1）,∫xsin（x^2+1）dx 2.已知∫f（x）dx=xe^x+才,则f（x
-
如题.1 ∫cos²x-sin²x／cos²xsin²x dx 2 ∫1＋cos
-
几道微积分题∫xsin(x^2)cos3(x^2)dx∫(cscx)^5dx
-
d2y/dx2+4y=xsin2x.
-
求定积分∫xsin^2x dx [-π/2,π/2]