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- LOAD("int_misc.mth")
-
- ;Removing discontinuities at x=-1 and x=1
- MAKECONT(x^2*SIGN(x^2-1),x,[-1,1])
-
- ;Fail to remove the discontinuity, because of the lack of side limits
- MAKECONT(SIN(1/x),x,[0])
-
-
- ;DERIVE finds a discontinuous antiderivative
- INT(ASIN(2*x/(x*x+1)),x)
-
- ;The correct (continuous) antiderivative
- CONT_INT(ASIN(2*x/(x*x+1)),x,[-1,1])
-
- ;The result is wrong!
- INT(ASIN(2*x/(x*x+1)),x,0,SQRT(3))
-
- ;Now it is OK
- CONT_INT(ASIN(2*x/(x*x+1)),x,[-1,1],0,SQRT(3))
-
- ;Fail to compute the antiderivative
- INT(EXP(ASIN(x)),x)
-
- ;Trying with INTSUBST
- INTSUBST(EXP(ASIN(x)),x,ASIN(x))
-
- ;Fail to compute the antiderivative
- INT((x+1)/(SQRT(x*x+x+1)+2*x+3),x)
-
- ;Using the first Euler substitution
- INTSUBST((x+1)/(SQRT(x*x+x+1)+2*x+3),x,x*x+x+1="euler1")
-
- ;Giving both the direct and inverse substitutions
- INTSUBST((x+1)/(SQRT(x*x+x+1)+2*x+3),x,SQRT(x*x+x+1)=x+t,t,x*x+x+1=(x+t)^2)
-
- ;Using the second Euler substitution
- INTSUBST((x+1)/(SQRT(x*x+x+1)+2*x+3),x,x*x+x+1="euler2")
-
- ;Using the third Euler substitution
- INTSUBST(SQRT(x*x-5*x+4),x,x*x-5*x+4="euler3")
-
- ;A definite integral
- INTSUBST((x+1)/(SQRT(x*x+x-6)+2*x+3),[x,2,3],x*x+x-6="euler1")
-
- ;An improper integral
- INTSUBST(1/(x*SQRT(x^2+x+1) + 2*x + 1), [x, 0, inf], x^2 + x + 1 = "euler1")
-
- ;INTSUBST cannot automatically compute the inverse substituion
- INTSUBST((x+1)/(SQRT(x*x+x+1)+2*x+3),x,SQRT(x*x+x+1)=x+t,t)
-
- ;Thank you!
- "":='~
- [" Ok, end of the demo! ";~
- " ";~
- "Author: Valeriu Anisiu, anisiu@math.ubbcluj.ro ";~
- "address: Faculty of Mathematics and Computer Science";~
- " Babes-Bolyai University ";~
- " Kogalniceanu 1 Street ";~
- " 3400 Cluj-Napoca, Romania "]
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