∫tanx/x^2dx

Question

∫tanx/x^2 dx

Answer 1

设x=arctant
即原式可变为
∫t²arctantd(arctant)
=∫t^4/(t²+1)arctantdt
=∫t^4/(t²+1)arbtantdt
=∫[t²-1+1/(t²+1)]arctantdt①
而∫1/(t²+1)arctantdt
=∫arctantd(arctant)
=1/2(arbtant)²+c
∫(t²-1)arctantdt
=∫arctantd(1/3t³-t)
=(1/3t³-t)arctant-∫(1/3t³-t)d(arctant)
其中∫(1/3t³-t)d(arctant)
=1/3∫(t³-3t)/(t²+1)dt
=1/3∫[t-4t/(t²+1)]dt
=1/3[1/2t²-2ln(t²+1)]+c
由此可得①的积分
再将t=tanx回代可得结果.