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∫xarctanxdx/√1+x二次幂
=∫arctanxd√(1+x^2)
=arctanx√(1+x^2)-∫1/(1+x^2)*√(1+x^2)dx
=arctanx√(1+x^2)-∫1/√(1+x^2)dx
对∫1/√(1+x^2)dx
令x=tant,dx=(sect)^2dt,√(1+x^2)=sect
∫1/√(1+x^2)dx=∫sectdt=ln|sect+tant|+c
=ln|x+√(1+x^2)|+c
∫xarctanxdx/√1+x二次幂
=arctanx√(1+x^2)-∫1/√(1+x^2)dx
=arctanx√(1+x^2)-ln|x+√(1+x^2)|+c
=∫arctanxd√(1+x^2)
=arctanx√(1+x^2)-∫1/(1+x^2)*√(1+x^2)dx
=arctanx√(1+x^2)-∫1/√(1+x^2)dx
对∫1/√(1+x^2)dx
令x=tant,dx=(sect)^2dt,√(1+x^2)=sect
∫1/√(1+x^2)dx=∫sectdt=ln|sect+tant|+c
=ln|x+√(1+x^2)|+c
∫xarctanxdx/√1+x二次幂
=arctanx√(1+x^2)-∫1/√(1+x^2)dx
=arctanx√(1+x^2)-ln|x+√(1+x^2)|+c
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