1/x^2*根号1+x^2 不定积分

顺便这个公式不会用，请教过程是怎么的，谢谢

∫ dx/[x²√(x²+a²)]
令x=a*tany，dx=a*sec²y dy
√(x²+a²)=√(a²*tan²y+a²)=a*secy
原式=∫ a*sec²y/(a²*tan²y*a*secy) dy
=(1/a²)∫secy/tan²y dy
=(1/a²)∫1/cosy*cos²y/sin²y dy
=(1/a²)∫cscycoty dy
=(-1/a²)cscy + C
由√(x²+a²)=√(a²*tan²y+a²)=a*secy
secy=√(x²+a²)/a
cosy=a/√(x²+a²)，cos²y=a²/(x²+a²)
cscy=1/siny=1/√(1-cos²y)=1/√[1-a²/(x²+a²)]=1/√[x²/(x²+a²)]=√(x²+a²)/x
∴原式=(-1/a²)*√(x²+a²)/x + C
=-√(x²+a²)/(a²x) + C

令x=tant,t∈(-π/2,π/2),则√(1+x²)=sect, dx=sec²tdt ∫√(1+x²) dx =∫sec³t dt =∫sect d(tant) =sect*tant-∫tant d(sect) =sect*tant-∫tan²t*sectdt =sect*tant-∫(sec²t-1)*sectdt =sect*tant-∫sec³tdt+∫sectdt ∴∫sec^3tdt=(1/2)(sect*tant+∫sectdt) =(1/2)(sect*tant+ln|sect+tant|)+c ∴原式=(1/2)[x*√(x^2+1)+ln|√(x^2+1)+x|]+c c为任意常数