用换元法求xarctanxdx的不定积分

Question

用换元法求xarctanxdx的不定积分

Answer 1

解：∫xarctanxdx=∫arctanxd(x²/2)=x²arctanx/2-(1/2)∫x²dx/(1+x²) (应用分部积分法)
=x²arctanx/2-(1/2)∫[1-1/(1+x²)]dx
=x²arctanx/2-(x-arctanx)/2+C (C是任意常数)
=(x²arctanx+arctanx-x)/2+C。

Answer 2

∫xarctanxdx=1/2∫arctanxdx^2
=x^2arctanx/2-1/2∫x^2darctanx
=x^2arctanx/2-1/2∫x^2/(1+x^2)dx
=x^2arctanx/2-1/2∫1-1/(1+x^2)dx
=x^2arctanx/2-x/2+arctanx+C

Answer 3

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