Question

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

不要吝啬步骤啊，不要跳步啊，大大们

Answer 1

可以考虑换元法，答案如图所示

Answer 2

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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