求积分dx/根号下[x+(根号x)]
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求积分dx/根号下[x+(根号x)]
求积分dx/根号下[x+(根号x)]
求积分dx/根号下[x+(根号x)]
令u=√x,则du=dx/(2√x)
∫dx/√(x+√x)
=2∫ u/√(u²+u) du
=2∫ u/√[(u+1/2)²-1/4] du
=2∫ (1/2·sect-1/2)/√[1/4·sect-1/4]·1/2·tant·sect dt 【令1/2·sect=u+1/2,du=1/2·tant·sectdt】
=∫(sec²t-sect) dt
=∫sec²tdt-∫sectdt
=tant-ln|tant+sect|+C
=2√(x+√x)-ln|2√x+2√(x+√x)+1|+C
令x=(tant)^4
4∫(sint)^2/(cost)^3dt
4∫(sint)^2/(cost)^4d(sint)
令sint=m
4∫m^2/(1+m^2)^2dm
令m=1/n
-4∫1/(n^2-1)^2dn
裂项
-∫[1/(n-1)-1/(n+1)]^2dn
得到
-∫1/(n-1)^2dn-∫1/(n...
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令x=(tant)^4
4∫(sint)^2/(cost)^3dt
4∫(sint)^2/(cost)^4d(sint)
令sint=m
4∫m^2/(1+m^2)^2dm
令m=1/n
-4∫1/(n^2-1)^2dn
裂项
-∫[1/(n-1)-1/(n+1)]^2dn
得到
-∫1/(n-1)^2dn-∫1/(n+1)^2dn+2∫1/(n-1)(n+1)dn
积分前面两项最后一项再
1/(n-1)+1/(n+1)+∫1/(n-1)-1/(n+1)dn
1/(n-1)+1/(n+1)+ln(n-1)-ln(n+1)+C
收起