∫1/根号下(1-25x^2)dx ∫1/(1+9x^2)dx 利用换元积分法求上面的不定积分~要过程~谢谢

解
∫1/√(1-25x²)dx
=1/5∫1/√[1-(5x)²]d(5x)
=1/5∫1/√(1-u²)du
=1/5acrsinu+C
=1/5arasin(5x)+C

∫1/(1+9x²)dx
=1/3∫1/[1+(3x)²]d(3x)
=1/3∫1/(1+u²)du
=1/3arctanu+C
=1/3arctan(3x)+C本回答被提问者采纳