Integrate 1/[x(x^2+1)^1/2]?

Let x = tanθ
dx = sec²θ dθ, cotθ = 1/x, cosecθ = [√(x² + 1)]/x

∫{1 / [x (x^2 + 1)^(1/2)]} dx
= ∫ {1 / [tanθ (tan²θ + 1)^(1/2)]} sec²θ dθ
= ∫ {1 / [tanθ secθ]} sec²θ dθ
= ∫ [secθ / tanθ] dθ
= ∫ [(1/cosθ) / (sinθ/cosθ)] dθ
= ∫(1/sinθ) dθ
= ∫cosecθ dθ
= ln|cosecθ - cotθ| + C
= ln|{[√(x² + 1)]/x} - (1/x)| + C

　
u = √(x²+1)
u² = x² + 1
2u du = 2x dx
u du = x dx

∫ 1/(x√(x²+1)) dx
= ∫ x/(x²√(x²+1)) dx
= ∫ u/((u²−1)u) du
= ∫ 1/(u²−1) du
= ∫ 1/2 (1/(u−1) − 1/(u+1)) du
= 1/2 (ln|u−1| − ln|u+1|) + C
= 1/2 (ln|√(x²+1)−1| − ln|√(x²+1)+1|) + C

= 1/2 ln|(√(x²+1)−1)/(√(x²+1)+1)| + C
= 1/2 ln|(√(x²+1)−1)(√(x²+1)−1)/ ((√(x²+1)+1)(√(x²+1)−1))| + C
= 1/2 ln|(√(x²+1)−1)²/((x²+1)−1)| + C
= 1/2 ln|(√(x²+1)−1)²/x²| + C
= ln|(√(x²+1)−1)/x| + C
= ln|√(x²+1)−1| − ln|x| + C