t²/(t²-2t-3)
=1+(2t+3)/(t-3)(t+1)
=1+(9/4)/(t-3)-(1/4)/(t+1)
所以原式=t+(9/4)ln|t-3|-(1/4)ln|t+1|+C
求∫t2/（t2-2t-3） dx
2个回答
相关问题
-
dx/dt=6t+2,dy/dt=（3t+1）sin（t^2）,求d^2y/dx^2
-
x=(t^2-2)\(t+1) y=(2t)\(t+1) 求dy\dx
-
参数方程求导x=f（t）-派y=f（e^3t-1） 求dy/dx t=0 d^2y/dX^2 t=0
-
x=t^2+t y=ln(1+t) 求dy/dx
-
x=2t+cost y=t+e^t 求dy/dx
-
已知x=2t+t^2,y=u(t),求得dy/dx=u'(t)/(2+2t),到这里都没问题,但是接下来求d^2y/dx
-
f(t)=∫[1,-1]|x/(1+x^2)-t|dx 求f(t)
-
matlab 参数方程x=1-t^2,y=t-t^3,画出图形,求导数d^2y/dx^2,并求所示点点二阶导数 t=√2
-
t*f(t)-∫f(x)dx(此处积分的下限为0,上限为t）=1/2*t2*f(t),求f(t).
-
设x=e^-t y=e^-2t 求dy/dx