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 Discussion Response Post to:
solve
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| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 May 2008 17:08:59 IST
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yup it is same as tan-1(2x+2/3).........just using pythagoras....
and now let 2x+2/3 = t
so dx = 3/2dt
now substitute in question.....
u will get 3/2integration tan -1tdt.....
now by using by parts method......
= 3/2[t tan -1 t -integration 1/1+t^2 * tdt
= 3/2[t tan-1 t - 1/2log(1+t^2)]......simplyfy t/1+t^2 ,,,,by assuming 1+t^2 = u and we get tdt = 1/2du........so the final answer is
3/2(2x+2/3)tan-1(2x+2/3) -3/4log(4x^2+8x+13/9)....
if i helped u plz rate me plzzzz
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if i helped u plzzzzz rate me,,,,,,, |
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