intigrate w.r.t x

Asked by sumahr | 20th Dec, 2009, 07:55: PM

Expert Answer:

f(x) = (x3+x)/(x4-9) = x(x2+1)/[(x2-3)(x2+3)]

Put x2+1 = y, and xdx = dy/2

Hence f(x)dx = (x3+x)dx/(x4-9) = x(x2+1)dx/[(x2-3)(x2+3)]

= ydy/[2(y-4)(y+2)]

Partial fractionating,

= (dy/2)[2/(y-4) + 1/(y+2)]

(x3+x)dx/(x4-9) = (dy/2)[2/(y-4) + 1/(y+2)] = (1/2)[2log(y-4) + log(y+2)] + c

=(1/2)log [(x2-3)2(x2+3)] + c =log [(x2-3)2(x2+3)] + c

Regards,

Team,

TopperLearning.

Answered by  | 20th Dec, 2009, 11:44: PM

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