∫ln(1+x^2)dx详细过程~

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∫ln(1+x^2)dx详细过程~∫ln(1+x^2)dx详细过程~∫ln(1+x^2)dx详细过程~用分部积分法,即∫udv=uv-∫vdu∫ln(1+x^2)dx=x*ln(1+x^2)-∫x*d

∫ln(1+x^2)dx详细过程~
∫ln(1+x^2)dx
详细过程~

∫ln(1+x^2)dx详细过程~
用分部积分法,即∫udv=uv-∫vdu
∫ln(1+x^2)dx
=x*ln(1+x^2)-∫x*d[ln(1+x^2)]
=x*ln(1+x^2)-∫x*[2x/(1+x^2)]dx
=x*ln(1+x^2)-2∫[x^2/(1+x^2)]dx
=x*ln(1+x^2)-2∫[(1+x^2)-1]/(1+x^2)dx
=x*ln(1+x^2)-2∫[1-(1/1+x^2)]dx
=x*ln(1+x^2)-2∫dx+2∫[1/(1+x^2)]dx
=x*ln(1+x^2)-2x+2arctanx+C.

原式=xln(1+x^2)-∫xdln(1+x^2)
=xln(1+x^2)-∫x/(1+x^2)*2xdx
=xln(1+x^2)-2∫(1+x^2-1)/(1+x^2)dx
=xln(1+x^2)-2∫[1-1/(1+x^2)]dx
=xln(1+x^2)-2x+2arctanx+C