∫(1-x-x∧2)／(x^2+1)^2dx给我说下吧.

Question

∫(1-x-x∧2)／(x^2+1)^2dx
给我说下吧.

Answer 1

∫ (1-x-x²)/(x²+1)² dx
=∫ (2-x-x²-1)/(x²+1)² dx
=2∫ 1/(x²+1)² dx - ∫ x/(x²+1)² dx - ∫ (x²+1)/(x²+1)² dx
=2∫ 1/(x²+1)² dx - (1/2)∫ 1/(x²+1)² dx² - ∫ 1/(x²+1) dx
=2∫ 1/(x²+1)² dx + (1/2)[1/(x²+1)] - arctanx
第一个积分换元,令x=tanu,则dx=sec²udu
=2∫ sec²u/(secu)^4 du + (1/2)[1/(x²+1)] - arctanx
=2∫ cos²u du + (1/2)[1/(x²+1)] - arctanx
=∫ (1+cos2u) du + (1/2)[1/(x²+1)] - arctanx
=u + (1/2)sin2u + (1/2)[1/(x²+1)] - arctanx + C
=u + sinucosu + (1/2)[1/(x²+1)] - arctanx + C
=arctanx + x/(1+x²) + (1/2)[1/(x²+1)] - arctanx + C
=(x+1/2)/(x²+1) + C
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