Question

(x的平方)乘以arccot x 的不定积分怎么求?

Answer 1

∫x^2(arccotx)dx

=(1/3)∫(arccotx)dx^3

=(1/3)x^3 arccotx -(1/3)∫x^3darccotx darccotx=darctan(1/x)=(-1/x^2)[1/(1+1/x^2)]=-1/(1+x^2)

=(1/3)x^3 arccotx+(1/3)∫x^3dx/(1+x^2)

=(1/3)x^3 arccotx +(1/12)∫dx^4/(1+x^2)

=(1/3)x^3 arccotx +(1/6)∫x^2dx^2/(1+x^2)

=(1/3)x^3 arccotx +(1/6)∫dx^2-(1/6)∫dx^2/(1+x^2)

=(1/3)x^3 arccotx +(1/6)x^2-(1/6)ln(1+x^2)+C