题目
[根号下(x^2-9)]/x的不定积分(换元法)
提问时间:2021-03-22
答案
令√(x^2-9)=u,则:x^2=u^2+9,∴d(x^2)=2udu.
∴∫[√(x^2-9)/x]dx
=(1/2)∫[2x√(x^2-9)/x^2]dx
=(1/2)∫[√(x^2-9)/x^2]d(x^2)
=(1/2)∫[u/(u^2+9)]·2udu
=∫{[(u^2+9)-9]/(u^2+9)}du
=∫du-9∫[1/(u^2+9)]du
=u-9∫{1/[9(u/3)^2+9]}du
=u-3∫{1/[(u/3)^2+1]}d(u/3)
=u-3arctan(u/3)+C
=√(x^2-9)-3arctan[(1/3)√(x^2-9)]+C.
∴∫[√(x^2-9)/x]dx
=(1/2)∫[2x√(x^2-9)/x^2]dx
=(1/2)∫[√(x^2-9)/x^2]d(x^2)
=(1/2)∫[u/(u^2+9)]·2udu
=∫{[(u^2+9)-9]/(u^2+9)}du
=∫du-9∫[1/(u^2+9)]du
=u-9∫{1/[9(u/3)^2+9]}du
=u-3∫{1/[(u/3)^2+1]}d(u/3)
=u-3arctan(u/3)+C
=√(x^2-9)-3arctan[(1/3)√(x^2-9)]+C.
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