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∫ln(1+x^2)*xdx
人气:298 ℃ 时间:2020-05-24 06:31:34
解答
原式=1/2∫ln(1+x²)d(1+x²)
=1/2(1+x²)ln(1+x²)-1/2∫(1+x²)dln(1+x²)
=1/2(1+x²)ln(1+x²)-1/2∫(1+x²)*1/(1+x²)*2xdx
=1/2(1+x²)ln(1+x²)-1/2∫2xdx
=1/2(1+x²)ln(1+x²)-x²/2+C
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