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Definite integral of

0 votes

[(x+1)]/[(x^2+2x-3)] from 0 to 1? Help!?

asked Dec 27, 2012 in CALCULUS by linda Scholar

1 Answer

0 votes

     1


= ʃx+1 / x^2+2x-3 dx                 ( multiple and divise with 2)

   0

     1

 
= ʃ  2(x+1) / 2(x^2+2x-3) dx

  0

=1/2(  ʃ 2x+2 / x^2+2x-3 dx)           ( ʃƒ1 (x) / f(x) = ln( f(x) )

                       1

=1/2[ln(x^2+2x-3)]

                       0

=1/2[(ln(1+2-3)-ln(-3)]

=1/2[ln(0)-ln(-3)]

=1/2[1-ln(-3)]

=1/2(1-0.477)

=1/2(0.53)

=0.53/2

=0.26

 1

ʃ(x+1) /(x^2+2x-3) dx = 0.26

0

answered Dec 27, 2012 by ricky Pupil

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