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求 x^4/x^4+5x^2+4 的不定积分
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解:
∫[x⁴/(x⁴+5x²+4)]dx
=⅓∫[x⁴[1/(x²+1) -1/(x²+4)]dx
=⅓∫[x⁴/(x²+1)]dx -⅓∫[x⁴/(x²+4)]dx
=⅓∫[(x⁴-1+1)/(x²+1)]dx-⅓∫[(x⁴-16+16)/(x²+4)]dx
=⅓∫[(x²-1)+ 1/(x²+1)]dx-⅓∫[(x²-4) +16/(x²+4)]dx
=⅓(⅓x³-x +arctanx)-⅓∫(x²-4)dx-(8/3)∫1/[(x/2)²+1)] d(x/2)
=(1/9)x³ -⅓x+⅓arctanx -⅓(⅓x³-4x) -(8/3)arctan(x/2) +C
=(1/9)x³ -⅓x+⅓arctanx -(1/9)x³+(4/3)x -(8/3)arctan(x/2) +C
=x+⅓arctanx -(8/3)arctan(x/2) +C
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