丫旦
幼苗
共回答了17个问题采纳率:100% 举报
∫ t²/(1 + t) dt
= ∫ t(t + 1 - 1)/(1 + t) dt
= ∫ t dt - ∫ t/(1 + t) dt
= t²/2 - ∫ (t + 1 - 1)/(1 + t) dt
= t²/2 - ∫ dt + ∫ dt/(1 + t)
= t²/2 - t + ln|1 + t| + C
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∫ x/(1 - x²) dx
= ∫ d(x²/2)/(1 - x²)
= (-1/2)∫ d(1 - x²)/(1 - x²)
= (-1/2)ln|1 - x²| + C
1年前
追问
5
wlwy2004
举报
= ∫ d(x²/2)/(1 - x²) = (-1/2)∫ d(1 - x²)/(1 - x²) 这2步怎么相等的,我不会算
举报
丫旦
d(x²/2) = (1/2)d(x²) = (1/2)d[(-x²)/(-1)] = (1/2)(-1)d(-x²) = (-1/2)d(-x²) = (-1/2)d(-x² + 1) = (-1/2)d(1 - x²) 逐步凑微分