1樓：匿名使用者
∫arctanxdx/[x^zhi2(1+x^dao2)]
=∫專arctanxdx/x^屬2 -∫arctanxdx/(1+x^2)
=∫arctanxd(-1/x)-∫arctanxdarctanx
=-(arctanx)/x +∫(1/x)darctanx-(arctanx)^2/2
=-(arctanx)/x-(arctanx)^2/2+∫dx/[x(1+x^2)]
其中 ∫dx/[x(1+x^2)]=∫[(1+x^2)-x^2]dx/[x(1+x^2)]=∫dx/x-∫xdx/(1+x^2)=lnx-(1/2)ln(1+x^2)+c
原式=-(arctanx)/x-(arctanx)^2/2+lnx-(1/2)ln(1+x^2)+c
2樓：我才是無名小將
^1/(x2(1+x2))=1/x^權2-1/(x^2+1)
1/x *1/(x^2+1)=1/x-x/(x^2+1)
s[arctanx/(x2(1+x2))]dx
=s[arctanx*(1/x^2-1/(x^2+1))dx
=sarctanx*1/x^2dx-sarctanx*1/(x^2+1)dx
=-sarctanx d(1/x)-sarctanx darctanx
=-arctanx *1/x+s1/x *darctanx -(arctanx )^2 *1/2
=-arctanx *1/x-(arctanx )^2 *1/2+s (1/x *1/(x^2+1)*dx
=-arctanx *1/x-(arctanx )^2 *1/2+s(1/x)*dx-s(x/(x^2+1)dx
=-arctanx *1/x-(arctanx )^2 *1/2+lnx-1/2*s1/(x^2+1) d(x^2+1)
==-arctanx *1/x-(arctanx )^2 *1/2+lnx-1/2*ln(x^2+1)+c
求(arctanx)/(x^2*(1+x^2))的不定積分
3樓：匿名使用者
^^|∫
zhiarctanxdx/(x^dao2(1+x^2)=∫版arctanxdx/x^2-∫arctanxdx/(1+x^2)
=-arctanx/x+∫dx/x(1+x^2)-(1/2)(arctanx)^2
=-arctanx/x+(1/2)ln[|權x^2|/|1+x^2|]-(1/2)(arctanx)^2+c
4樓：匿名使用者
|∫62616964757a686964616fe4b893e5b19e31333330336330 tan⁻¹x/[x²(1 + x²)] dx
= ∫ tan⁻¹x d(- 1/x - tan⁻¹x)
= tan⁻¹x · (- 1/x - tan⁻¹x) - ∫ (- 1/x - tan⁻¹x) d(tan⁻¹x)
= - (tan⁻¹x)/x - (tan⁻¹x)² + ∫ (1/x + tan⁻¹x)/(1 + x²) dx
= - (tan⁻¹x)/x - (tan⁻¹x)² + ∫ [(1 + x²) - x²]/[x(1 + x²)] + ∫ tan⁻¹x/(1 + x²) dx
= - (tan⁻¹x)/x - (tan⁻¹x)² + ∫ 1/x dx - ∫ x/(1 + x²) dx + ∫ tan⁻¹x d(tan⁻¹x)
= - (tan⁻¹x)/x - (tan⁻¹x)² + ln|x| - (1/2)ln(1 + x²) + (1/2)(tan⁻¹x)² + c
= - (1/2)ln(1 + x²) - (1/2)(tan⁻¹x)² - (tan⁻¹x)/x + ln|x| + c
求arctanx/1+x2的不定積分，急、急。謝謝咯
5樓：帥哥靚姐
(arctanx)'
=1/(tanx)'
=1/sec²x
=1/(tan²x+1)
=1/(x²+1)
∫arctanx/(1+x²)
=∫(arctanx)dx
=((arctanx)²/2)+c
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