积分∫x/[(x^2+1)(x^2+4)]dx

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积分∫x/[(x^2+1)(x^2+4)]dx
积分∫x/[(x^2+1)(x^2+4)]dx

积分∫x/[(x^2+1)(x^2+4)]dx
∫x/[(x^2+1)(x^2+4)]dx
=1/3∫x[1/(x^2+1) - 1/(x^2+4)]dx
=1/3[∫x/(x^2+1)dx-∫x/(x^2+4)dx]
=1/3[1/2∫1/[(x^2+1)]d(x^2+1)-1/2∫1/[(x^2+4)d(x^2+4)]]
=1/3[1/2ln(x^2+1)-1/2ln(x^2+4)+C]
=1/6ln[(x^2+1)/(x^2+4)]+C

原式=1/2∫dx²/(x²+1)(x²+4)
=1/2∫1/3·[1/(x²+1)-1/(x²+4)]dx²
=1/6∫dx²/(x²+1)-1/6∫dx²/(x²+4)
=1/6·ln(x²+1)-1/6·ln(x²+4)+C

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