Evaluate
Asked by harshil92
| 19th Feb, 2010,
03:10: PM
(x2-x+1) / (2x2+x+2) = (2x2-2x+2) /[2(2x2+x+2)] ...........multiply and divide by 2
= (2x2+x+2 - 3x) /[2(2x2+x+2)]
=(1/2)[1 - 3x/(2x2+x+2)]
= (1/2)[1 - (3/4)(4x+1-1)/(2x2+x+2)] ..........3x = (3/4)(4x+1-1)
= (1/2)[1 - (3/4)(4x+1)/(2x2+x+2) +(3/4)/(2x2+x+2)]
(x2-x+1) dx/ (2x2+x+2) =
(1/2)[1 - (3/4)(4x+1)/(2x2+x+2) +(3/4)/(2x2+x+2)]dx =
(1/2){dx - (3/4)
(4x+1)dx/(2x2+x+2) + (3/4)
dx/(2x2+x+2)} =
(1/2){x - (3/4) ln (2x2+x+2) + (3/4)dx/(
2x+1/2
2)2 + (
(15/8))2}
= (1/2){x - (3/4) ln (2x2+x+2) + (3/4)((8/15))(
2)tan-1((
2x + 1/2
2)/
(15/8))} + c
= (1/2){x - (3/4) ln (2x2+x+2) + (3/5) tan-1((4x + 1 )/(
15))} + c
Regards,
Team,
TopperLearning.
Answered by
| 22nd Feb, 2010,
09:14: AM
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