∫xe^arctanx⼀(1+x^2)^3⼀2dx 次方 求解啊，知道是换元法，求详解啊

你好

令arctanx=t，则x=tant，dx=sec²tdt
∫xe^arctanx/(1+x^2)^3/2dx
=∫tante^t/(1+tan^2;t)^3/2*sec²tdt
=∫tante^t/sec ³t sec ²tdt
=∫tante^t/sectdt
=∫sinte^tdt (1)
=-∫e^tdcost
=-coste^t+∫coste^tdt
=-coste^t+sinte^t-∫sinte^tdt (2)
由 (1)(2)得
∫sinte^tdt =1/2( sinte^t-coste^t ) +C
=1/2( sint-cost)e^t +C
=1/2cost(tant-1)e^t +C
=1/2*1/√(tan²t+1)*(tant-1)e^t +C
=1/2*1/√(x²+1)*(x-1)e^arctanx+C
=√(x²+1)*(x-1)e^arctanx/(x²+1)+C
即
∫xe^arctanx/(1+x^2)^3/2dx
=√(x²+1)*(x-1)e^arctanx/(x²+1)+C

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