Question

求定积分。。。

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Answer 1

(6)
∫(0->1) 3x^3/(1+x^2) dx
=∫(0->1) [3x - 3x/(1+x^2) ]dx
= [(3/2)x^2 - (3/2)ln|1+x^2| ]|(0->1)
=3/2 - (3/2)ln2
(12)
∫ x^5/√(1-x^2) dx

=-∫ x^4 d√(1-x^2)
=-x^4. √(1-x^2) + 4∫ x^3 .√(1-x^2) dx
=-x^4. √(1-x^2) -(4/3)∫ x^2 . d(1-x^2)^(3/2)
=-x^4. √(1-x^2) -(4/3)x^2 (1-x^2)^(3/2) +(8/3)∫ x. (1-x^2)^(3/2) dx
=-x^4. √(1-x^2) -(4/3)x^2 (1-x^2)^(3/2) -(4/3)∫ (1-x^2)^(3/2) d(1-x^2)
=-x^4. √(1-x^2) -(4/3)x^2 (1-x^2)^(3/2) -(8/15)(1-x^2)^(5/2) +C