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求不定积分∫xln(1+x^2)dx
人气:361 ℃ 时间:2020-05-10 22:22:45
解答
∫xln(1+x^2)dx
=1/2∫ln(1+x^2)dx^2
=1/2∫ln(1+x^2)d(1+x^2)
=1/2(1+x^2)ln(1+x^2)-1/2∫(1+x^2)dln(1+x^2)
=1/2(1+x^2)ln(1+x^2)-1/2∫(1+x^2)*1/(1+x^2)d(1+x^2)
=1/2(1+x^2)ln(1+x^2)-1/2∫dx^2
=1/2(1+x^2)ln(1+x^2)-1/2x^2+C
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