anonymous
  • anonymous
how do i integrate (1+x^2)/(1-x^2) from 0 to 0.5
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Zarkon
  • Zarkon
expand (1+x^2)/(1-x^2)
TuringTest
  • TuringTest
I wrote\[\int{1\over1-x^2}dx-\int{x^2\over1-x^2}dx\]the first one is a known hyperbolic tangent integral isn't it? I didn't want to say because I wasn't sure
TuringTest
  • TuringTest
+ in between*

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TuringTest
  • TuringTest
the second is a trig sub for sure though...
TuringTest
  • TuringTest
or long division...
Zarkon
  • Zarkon
\[\frac{1+x^2}{1-x^2}=\frac{1}{x+1}-\frac{1}{x-1}-1\]
TuringTest
  • TuringTest
oh yeah, that thing they call partial fractions !! I guess it eluded me at the moment :(
anonymous
  • anonymous
another way i saw of doing it would be to write 2-(1-x^2)/ (1-x^2)
anonymous
  • anonymous
[2-(1-x^2)]/ (1-x^2)
TuringTest
  • TuringTest
still leads to partial fractions that way ... actually that is what I meant by long division
TuringTest
  • TuringTest
...sort of
anonymous
  • anonymous
yeah i guess. thanks for the help
amistre64
  • amistre64
1+2x^2+2x^4+2x^6 ..... ------------- 1-x^2 ) 1+x^2 (1-x^2) -------- 2x^2 (2x^2-2x^4) ------------ 2x^4 .......... \[\int 1+2x^2+2x^4+2x^6 ...\ dx\] \[x+\frac{2}{3}x^3+\frac{2}{5}x^5+\frac{2}{7}x^7 ...\]
amistre64
  • amistre64
\[\sum_{n=0}^{inf}\frac{2(0.5)^{2n+1}}{2n+1}\]

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