高等数学不定积分换元法
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高等数学不定积分换元法
∫[arctanx/x^2(1+x^2)]dx 用换元法怎么解
∫[arctanx/x^2(1+x^2)]dx 用换元法怎么解
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换元法与分部法结合
令t=arctanx,则
∫[arctanx/x^2(1+x^2)]dx
=∫t/[(tant)^2×(sect)^2]×(sect)^2 dt
=∫t×(cott)^2 dt
=∫t×(csct)^2 dt-∫t dt
=-∫t d(cott)-1/2×t^2
=-t×cott+∫cottdt-1/2×t^2
=-t×cott+ln|sint|-1/2×t^2+C
=-arctanx/x+ln|x|-1/2×ln(1+x^2)-1/2×(arctanx)^2+C
令t=arctanx,则
∫[arctanx/x^2(1+x^2)]dx
=∫t/[(tant)^2×(sect)^2]×(sect)^2 dt
=∫t×(cott)^2 dt
=∫t×(csct)^2 dt-∫t dt
=-∫t d(cott)-1/2×t^2
=-t×cott+∫cottdt-1/2×t^2
=-t×cott+ln|sint|-1/2×t^2+C
=-arctanx/x+ln|x|-1/2×ln(1+x^2)-1/2×(arctanx)^2+C