Question

来，算一下这道反常积分。

Answer 1

望采纳哦

追答

有个地方写错了

应该比较清楚，望采纳哦

追问

非常感谢，可惜还是错的。。

非常感谢，可惜还是错的。。

Answer 2

原式=-∫[1,+∞]arctanx d(1/x)
=-arctanx /x |[1,+∞] +∫[1,+∞]dx/x(1+x²)
=-lim[x→+∞]arctanx/x +arctan1 +∫[1,+∞][1/x - x/(1+x²)]dx
=-lim[x→+∞]1/(1+x²) +π/4 +∫[1,+∞]1/x dx - ½ ∫[1,+∞]1/(1+x²) d(1+x²)
=-lim[x→+∞]1/(1+x²) +π/4 +[lnx -½ln(1+x²)]|[1,+∞]
=π/4 +½ ln[x²/(1+x²)]|[1,+∞]
=π/4+½ln[1- 1/(1+x²)]|[1,+∞]
=π/4 +½ (ln1 - ln½)
=π/4 -½ln½
=π/4 +½ln2

Answer 3

如图

追答

如图

Answer 4

如图

追问

倒数第三步不够详细

倒数第三步不够详细

Answer 5

设 x = tanu， 则
I = ∫<π/4, π/2> u(secu)^2du/(tanu)^2 = ∫<π/4, π/2> udu/(sinu)^2
= -∫<π/4, π/2> udcotu = -[ucotu]<π/4, π/2> + ∫<π/4, π/2> cotudu
= π/4 + [lnsinu]<π/4, π/2> = π/4 + (1/2)ln2