Question

求不定积分：（3x^4+2x^3+4x^2+2x+5)/(x^2+1)

答案是：x^3+x^2+x+4arctanx+c

Answer 1

∫（3x^4+2x^3+4x^2+2x+5)/(x^2+1)dx
= ∫[3x^2(x^2+1)+2x(x^2+1)+(x^2+1)+4]/(x^2+1)dx
= ∫[3x^2+2x+1+4/(x^2+1)]dx
=x^3+x^2+x+4arctanx+C
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Answer 2

3x⁴ = 3x²[(x² + 1) - 1] = 3x²(x² + 1) - 3[(x² + 1) - 1] = 3x²(x² + 1) - 3(x² + 1) + 3
2x³ = 2x[(x² + 1) - 1] = 2x(x² + 1) - 2x
4x² = 4[(x² + 1) - 1] = 4(x² + 1) - 4
∴
3x⁴ + 2x³ + 4x² + 2x + 5
= [3x²(x² + 1) - 3(x² + 1) + 3] + [2x(x² + 1) - 2x] + [4(x² + 1) - 4] + 2x + 5
= (x² + 1)(3x² - 3 + 2x + 4) + (3 - 4 + 5)
= (x² + 1)(3x² + 2x + 1) + 4
∴
∫ [3x⁴ + 2x³ + 4x² + 2x + 5]/(x² + 1) dx
= [(x² + 1)(3x² + 2x + 1) + 4]/(x² + 1) dx
= ∫ (3x² + 2x +1) dx + 4∫ dx/(x² + 1)
= x³ + x² + x + 4arctan(x) + C