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求不定积分:∫ln(x^2+1)dx
人气:126 ℃ 时间:2020-03-22 05:28:48
解答
用分步积分
∫ln(x^2+1)dx
=xln(x^2+1)-∫2x^2/(x^2+1)dx
=xln(x^2+1)-∫(2x^2+2-2)/(x^2+1)dx
=xln(x^2+1)-∫[2-2/(x^2+1)]dx
=xln(x^2+1)-2x+2arctanx+C
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