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Integral Calculus
24 Oct 2007 17:45:38 IST
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24 Oct 2007 22:16:20 IST
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dx/ x^1/2 + x^1/3 dx/ x^1/3 ( 1+x^1/6 )Let x = t^6
dx = 6t^5dt
I = 6t^5dt / t^2(1+t)
6t^5dt / t^2(1+t) = 6 t^3 dt / 1+ t
t^3 dt / 1+ t = 6 (t^3 + 1 - 1)dt/ (1+t)
(t^3 + 1 - 1)dt/ (1+t) = 6 {(t^2 - t +1 - (1/ 1+t) } dt
{(t^2 - t +1 - (1/ 1+t) } dt = 6 [ t^3/3 - t^2/2 + t - log(1+t) ] + c
[ t^3/3 - t^2/2 + t - log(1+t) ] + c = 2t^3 - 3t^2 + 6t - 6log(1+t) + c
= 2
x - 3x^1/3 + 6x^1/6 - 6log(1+x1/6) + c
x - 3x^1/3 + 6x^1/6 - 6log(1+x1/6) + cFree Sign Up!
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dx = 6t^5dt
on substituting we get
I = 6t^5 dt
------------
t^3 + t^2
this can be solved by using partial fractions