what are the steps of integration ∫x^2 / (1+x^2) dx ?
the answer is x-tan^(-1)(x) + C
i can't figure out T^T
THX
integration ∫x^2 / (1+x^2) dx
2014-07-03 7:52 am
回答 (2)
2014-07-03 8:39 am
✔ 最佳答案
∫ (x² / (1+ x²) dx= ∫ {[(1 + x²) -1] / (1 + x²)} dx
= ∫ {1 - [1 / (1 + x²)] dx
= ∫ dx - ∫[1 / (1 + x²)] dx
= x - tan⁻¹x +C
參考: 土扁
2014-07-03 9:15 am
Other method：
∫ x^2 / (1+x^2) dx
Let x=tanθ
dx=sec^2 θ dθ
∫ x^2 / (1+x^2) dx
=∫ tan^2 θ sec^2 θ / (1 + tan^2 θ) dθ
=∫ tan^2 θ sec^2 θ/ sec^2 θ dθ
=∫ tan^2 θ dθ
=∫ (sec^2 θ - 1) dθ
=tanθ - θ +C
=x - tan^(-1)(x) + C
∫ x^2 / (1+x^2) dx
Let x=tanθ
dx=sec^2 θ dθ
∫ x^2 / (1+x^2) dx
=∫ tan^2 θ sec^2 θ / (1 + tan^2 θ) dθ
=∫ tan^2 θ sec^2 θ/ sec^2 θ dθ
=∫ tan^2 θ dθ
=∫ (sec^2 θ - 1) dθ
=tanθ - θ +C
=x - tan^(-1)(x) + C
收錄日期: 2021-04-15 15:57:06
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