Question

求不定积分(x^3+x−1)/(x^2+2)^2

Answer 1

被积函数 f(x) = (x^3+x−1)/(x^2+2)^2 = (x^3+2x-x−1)/(x^2+2)^2
= x/(x^2+2) - x/(x^2+2)^2 − 1/(x^2+2)^2
I1 = ∫xdx/(x^2+2) = (1/2)∫d(x^2+2)/(x^2+2) = (1/2)ln(x^2+2) + C1,
I2 = ∫xdx/(x^2+2)^2 = (1/2)∫d(x^2+2)/(x^2+2)^2 = -(1/2)/(x^2+2) + C2
I3 = ∫ dx/(x^2+2)^2 (令 x = √2tant)
= ∫ dx/(x^2+2)^2 = (√2/4)∫ dt/(sect)^2 = (√2/4)∫ (cost)^2dt
= (√2/8)∫ (1-cos2t)dt = (√2/8)[t-(1/2)sin2t] + C3
= (√2/8)[t-sintcost] + C3 = (√2/8)[arctan(x/√2)-√2x/(x^2+2)] + C3
I = I1 - I2 - I3 = (1/2)ln(x^2+2) + (1/2)/(x^2+2) - (√2/8)[arctan(x/√2)-√2x/(x^2+2)] + C

Answer 2

∫x³/√(1+x²) dx =∫ x² d(√(1+x²) ) =x²√(1+x²) - ∫ √(1+x²) d(x²) =x²√(1+x²) -2/3 *(1+x²)^(3/2) +C