integration
2011-03-09 4:43 am
It is given that x^2+2x+4 = [(x+1)^2]+3Hence show that ∫dx/(x^2+2x+4) = √3π/18
回答 (3)
2011-03-10 12:28 am
✔ 最佳答案
Is it a definite integral?2011-03-09 16:28:08 補充：
As follows:
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2011-03-09 5:43 am
yes...................
2011-03-08 21:46:57 補充：
sorry .... the upper limit is 2
lower limit is 0
2011-03-08 21:46:57 補充：
sorry .... the upper limit is 2
lower limit is 0
2011-03-09 5:35 am
∫dx/(x^2+2x+4) = ∫dx/ [(x+1)²+3]
= ∫d(x+1)/[(x+1)²+3]
Put x+1 = √3tanA
= ∫d( √3tanA)/( √3tan²A+3)
=√3 ∫ sec²A dA / [3(tan²A+1)]
= √3/3 ∫dA
=√3/3 A + C
= √3/3 tan^-1 ( x+1 /√3) + C
When the funtion is integrated from -1 to 0,
then the ans. is √3π/18
= ∫d(x+1)/[(x+1)²+3]
Put x+1 = √3tanA
= ∫d( √3tanA)/( √3tan²A+3)
=√3 ∫ sec²A dA / [3(tan²A+1)]
= √3/3 ∫dA
=√3/3 A + C
= √3/3 tan^-1 ( x+1 /√3) + C
When the funtion is integrated from -1 to 0,
then the ans. is √3π/18
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