please solve it

Asked by www1028 | 12th Sep, 2010, 10:08: PM

Expert Answer:

∫dx/(2x - x2)3/2 =
∫dx/(1 - 1 + 2x - x2)3/2 =
∫dx/(1 - (x - 1)2 )3/2 =
∫dx/[(1 - (x - 1))(1 + (x - 1))]3/2 =
∫dx/[(2 - x )x]3/2 =
∫dx/[(2 - x )3/2x3/2] =
Using partial fractions,
1/[(2 - x )3/2x3/2] = A/x3/2 + B/(2 - x )3/2
Hence we find,
A = 1/23/2 = B
∫dx/[(2 - x )3/2x3/2] =
(1/23/2)∫dx/(2 - x )3/2 + dx/x3/2] =
(1/23/2){2/(2 - x )1/2 -  2/x1/2] + C =
{1/(2(2 - x ))1/2 -  1/(2x)1/2] + C
regards,
Team,
TopperLearning.

Answered by  | 13th Sep, 2010, 09:55: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.