不定积分∫√（1＋X^2）dX的解过程

需借助三角函数换元.
x = tany、dx = sec^2y dy
∫ √(1 + x^2) dx
= ∫ √(1 + tan^2y) * sec^2y dy
= ∫ sec^3y dy
= ∫ secy d(tany)
= secytany - ∫ tany d(secy)
= secytany - ∫ tany * (secytany dy)
= secytany - ∫ (sec^2y - 1) * secy dy
= secytany - ∫ sec^3y dy + ∫ secy dy
2∫ sec^3y dy = secytany + ∫ secy dy
∫ sec^3y dy = (1/2)secytany + (1/2)ln|secy + tany| + C
= (x/2)√(1 + x^2) + (1/2)ln|x + √(1 + x^2)| + C