求解不定积分，我要详细步骤

Question

∫t²/(t+1)dt

∫x/(1-x²)dx

Answer 1

∫ 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
___________________________
∫ x/(1 - x²) dx
= ∫ d(x²/2)/(1 - x²)
= (-1/2)∫ d(1 - x²)/(1 - x²)
= (-1/2)ln|1 - x²| + C追问

= ∫ d(x²/2)/(1 - x²)
= (-1/2)∫ d(1 - x²)/(1 - x²)
这2步怎么相等的，我不会算

追答

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