anonymous
  • anonymous
integral[(tan^(-1)x)/(1+x^2)]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
put -1/x=t and calculate dx/dt
anonymous
  • anonymous
hey...hang on...try integration by parts...
amistre64
  • amistre64
make x = tan(t)

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anonymous
  • anonymous
put (tan^(-1)x) = t 1/1+x^2 dx = dt integratl (t dt) = t^2/2
amistre64
  • amistre64
uzmas on the right track :)
amistre64
  • amistre64
tan(t) = x ; x = tan-^1(x) dt sec^2 = dx ; 1+tan^2(t) = sec^2 tan^-1(x) t sec^2(t) dt -------- --> -------------- --> t dt 1 + x^2 sec^2(t)
amistre64
  • amistre64
[S] t dt = (t^2)/2 substitute back for t tan^-1(x) -------- is what it would be if you see thru the typos above lol 2
amistre64
  • amistre64
+C :)
amistre64
  • amistre64
[ tan^-1(x) ]^2 ----------- comeon people......let me know when im wrong ;) 2
amistre64
  • amistre64
+C....ack!!
anonymous
  • anonymous
you r right, but after substitution we can differentiate both sides
anonymous
  • anonymous
d/dx(tan ^(-1)x) = 1/!+ x^2
anonymous
  • anonymous
oh 1/1 + x^2
amistre64
  • amistre64
tan^-1(x) --> 1/(1+x^2)
anonymous
  • anonymous
yes right
anonymous
  • anonymous
n directly we get Integral (tdt)
amistre64
  • amistre64
and the (...)^2 --> 2(....) 2(.....)/2 = (.......) = tan^-1(x)/(1+x^2) :) yay!!
anonymous
  • anonymous
??
amistre64
  • amistre64
lol..... it makes sense to me :)
anonymous
  • anonymous
:)
anonymous
  • anonymous
So what method was used? trig substitution?
amistre64
  • amistre64
I used trig sub...yeah
amistre64
  • amistre64
i tend to forget that dt != dx ... and that there is something I miss :)

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