Question

我书上的基本积分公式好像错了，你们看看

∫dx/√(1+x²)=arcsinx+C
这个是不是书上写错了？

Answer 1

不确定就再推导一下

求不定积分∫1/√(a²+x²)dx

令x=atant，tant=x/a，dx=datant=asec²tdt
x=asint/cost
x²=a²sin²t/cos²t
x²=a²(1-cos²t)/cos²t
x²=a²sec²t-a²
sec²t=(a²+x²)/a²
得sect=√(a²+x²)/a

∫1/√(a²+x²)dx
=∫sec²t/√(a²+a²tan²t)dt
=∫asec²t/√a²sec²tdt
=∫asec²t/asectdt
=∫sec²t/sectdt
=∫cost/cos²tdt
=∫1/cos²tdsint
=∫1/(1-sin²t)dsint
=-∫1/(sint+1)(sint-1)dsint
=-∫[1/(sint-1)-1/(sint+1)]/2dsint
=-[∫1/(sint-1)dsint-∫1/(sint+1)dsint]/2
=[∫1/(sint+1)d(sint+1)-∫1/(sint-1)d(sint-1)]/2
=(ln|sint+1|-ln|sint-1|)/2+C
=ln√|(sint+1)/(sint-1)|+C
=ln√|(sint+1)²/(sint+1)(sint-1)|+C
=ln√|(sint+1)²/(sin²t-1)|+C
=ln√|-(sint+1)²/cos²t|+C
=ln|(sint+1)/cost|+C
=ln|tant+1/cost|+C
=ln|sect+tant|+C
=ln|√(a²+x²)/a+x/a|+C
=ln|x+√(a²+x²)|-lna+C
=ln|x+√(a²+x²)|+C

Answer 2

是错了，应该是∫dx/√(1+x²)=ln|x+√(1+x²)|+C

Answer 3

好像是∫dx/√(1-x²)=arcsinx+C