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题目
∫ x 4 x 6 + 1 d x \int {{{{x^4}} \over {{x^6} + 1}}dx} ∫x6+1x4dx
来源
思考
在做积分题时，若遇到不会积分的，我们可以考虑先积简单（形式和原式一样）的积分：
∫ x 6 + 1 x 6 + 1 d x = x + C ( 1 ) \int {{{{x^6} + 1} \over {{x^6} + 1}}dx} = x + C(1) ∫x6+1x6+1dx=x+C(1)
∫ x 5 x 6 + 1 d x = 1 6 ln ( x 6 + 1 ) + C ( 2 ) \int {{{{x^5}} \over {{x^6} + 1}}dx} = {1 \over 6}\ln \left( {{x^6} + 1} \right) + C(2) ∫x6+1x5dx=61ln(x6+1)+C(2)
∫ x 2 x 6 + 1 d x = ∫ x 2 ( x 3 ) 2 + 1 d x = 1 3 arctan x 3 + C ( 3 ) \int {{{{x^2}} \over {{x^6} + 1}}dx} = \int {{{{x^2}} \over {{{\left( {{x^3}} \right)}^2} + 1}}dx} = {1 \over 3}\arctan {x^3} + C(3) ∫x6+1x2dx=∫(x3)2+1x2dx=31arctanx3+C(3)
至此，我们已经得到了三块积木： x 6 + 1 、 x 5 、 x 2 {{x^6} + 1}、x^5、x^2 x6+1、x5、x2，接着思考，分母我们可以看作立方和： x 6 + 1 = ( x 2 ) 3 + 1 = ( x 2 + 1 ) ( x 4 − x 2 + 1 ) {x^6} + 1 = {\left( {{x^2}} \right)^3} + 1 = \left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right) x6+1=(x2)3+1=(x2+1)(x4−x2+1)我们可以得到： ∫ x 4 − x 2 + 1 x 6 + 1 d x = ∫ 1 x 2 + 1 d x = arctan x + C ( 4 ) \int {{{{x^4} - {x^2} + 1} \over {{x^6} + 1}}dx} = \int {{1 \over {{x^2} + 1}}dx} = \arctan x + C(4) ∫x6+1x4−x2+1dx=∫x2+11dx=arctanx+C(4)我们知道 ∫ x 2 + 1 x 4 − x 2 + 1 d x = ∫ 1 + 1 x 2 x 2 − 1 + 1 x 2 d x = ∫ d ( x − 1 x ) ( x − 1 x ) 2 + 1 = arctan ( x − 1 x ) + C \int {{{{x^2} + 1} \over {{x^4} - {x^2} + 1}}dx} = \int {{{1 + {1 \over {{x^2}}}} \over {{x^2} - 1 + {1 \over {{x^2}}}}}dx} = \int {{{d\left( {x - {1 \over x}} \right)} \over {{{\left( {x - {1 \over x}} \right)}^2} + 1}} = \arctan \left( {x - {1 \over x}} \right)} + C ∫x4−x2+1x2+1dx=∫x2−1+x211+x21dx=∫(x−x1)2+1d(x−x1)=arctan(x−x1)+C ∫ x 2 − 1 x 4 − x 2 + 1 d x = ∫ 1 − 1 x 2 x 2 − 1 + 1 x 2 d x = ∫ d ( x + 1 x ) ( x + 1 x ) 2 − 3 = 1 2 3 ln ∣ x + 1 x − 3 x + 1 x + 3 ∣ + C \int {{{{x^2} - 1} \over {{x^4} - {x^2} + 1}}dx} = \int {{{1 - {1 \over {{x^2}}}} \over {{x^2} - 1 + {1 \over {{x^2}}}}}dx} = \int {{{d\left( {x + {1 \over x}} \right)} \over {{{\left( {x + {1 \over x}} \right)}^2} - 3}} = {1 \over {2\sqrt 3 }}\ln \left| {{{x + {1 \over x} - \sqrt 3 } \over {x + {1 \over x} + \sqrt 3 }}} \right|} + C ∫x4−x2+1x2−1dx=∫x2−1+x211−x21dx=∫(x+x1)2−3d(x+x1)=231ln x+x1+3x+x1−3 +C我们对上面两个式子分子分母同乘 x 2 + 1 x^2+1 x2+1，即可得到 ∫ x 4 + 2 x 2 + 1 x 6 + 1 d x = ∫ ( x 2 + 1 ) ( x 2 + 1 ) ( x 4 − x 2 + 1 ) ( x 2 + 1 ) d x = ∫ x 2 + 1 x 4 − x 2 + 1 d x = ∫ 1 + 1 x 2 x 2 − 1 + 1 x 2 d x = ∫ d ( x − 1 x ) ( x − 1 x ) 2 + 1 = arctan ( x − 1 x ) + C ( 5 ) \int {{{{x^4} + 2{x^2} + 1} \over {{x^6} + 1}}dx} = \int {{{\left( {{x^2} + 1} \right)\left( {{x^2} + 1} \right)} \over {\left( {{x^4} - {x^2} + 1} \right)\left( {{x^2} + 1} \right)}}dx} = \int {{{{x^2} + 1} \over {{x^4} - {x^2} + 1}}dx} = \int {{{1 + {1 \over {{x^2}}}} \over {{x^2} - 1 + {1 \over {{x^2}}}}}dx} = \int {{{d\left( {x - {1 \over x}} \right)} \over {{{\left( {x - {1 \over x}} \right)}^2} + 1}} = \arctan \left( {x - {1 \over x}} \right)} + C(5) ∫x6+1x4+2x2+1dx=∫(x4−x2+1)(x2+1)(x2+1)(x2+1)dx=∫x4−x2+1x2+1dx=∫x2−1+x211+x21dx=∫(x−x1)2+1d(x−x1)=arctan(x−x1)+C(5) ∫ x 4 − 1 x 6 + 1 d x = ∫ ( x 2 − 1 ) ( x 2 + 1 ) ( x 4 − x 2 + 1 ) ( x 2 + 1 ) d x = ∫ x 2 − 1 x 4 − x 2 + 1 d x = ∫ 1 − 1 x 2 x 2 − 1 + 1 x 2 d x = ∫ d ( x + 1 x ) ( x + 1 x ) 2 − 3 = 1 2 3 ln ∣ x + 1 x − 3 x + 1 x + 3 ∣ + C ( 6 ) \int {{{{x^4} - 1} \over {{x^6} + 1}}dx} = \int {{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)} \over {\left( {{x^4} - {x^2} + 1} \right)\left( {{x^2} + 1} \right)}}dx} = \int {{{{x^2} - 1} \over {{x^4} - {x^2} + 1}}dx} = \int {{{1 - {1 \over {{x^2}}}} \over {{x^2} - 1 + {1 \over {{x^2}}}}}dx} = \int {{{d\left( {x + {1 \over x}} \right)} \over {{{\left( {x + {1 \over x}} \right)}^2} - 3}} = {1 \over {2\sqrt 3 }}\ln \left| {{{x + {1 \over x} - \sqrt 3 } \over {x + {1 \over x} + \sqrt 3 }}} \right|} + C(6) ∫x6+1x4−1dx=∫(x4−x2+1)(x2+1)(x2−1)(x2+1)dx=∫x4−x2+1x2−1dx=∫x2−1+x211−x21dx=∫(x+x1)2−3d(x+x1)=231ln x+x1+3x+x1−3 +C(6)
至此我们一共得到了六块积木： x 6 + 1 、 x 5 、 x 2 、 x 4 − x 2 + 1 、 x 4 + 2 x 2 + 1 、 x 4 − 1 {{x^6} + 1}、x^5、x^2、x^4-x^2+1、{{x^4} + 2{x^2} + 1}、{{x^4} - 1} x6+1、x5、x2、x4−x2+1、x4+2x2+1、x4−1，我们接着搭积木，将这些积木组合为分子 x 4 x^4 x4， x 4 = 1 2 ( x 4 − 1 ) + 1 2 ( x 4 − x 2 + 1 ) + 1 2 x 2 {x^4} = {1 \over 2}\left( {{x^4} - 1} \right) + {1 \over 2}\left( {{x^4} - {x^2} + 1} \right) + {1 \over 2}{x^2} x4=21(x4−1)+21(x4−x2+1)+21x2这里我们只展示这一种，当然还有很多种组合方法，读者可以自行尝试。
证明
∫ x 4 x 6 + 1 d x = ∫ 1 2 ( x 4 − 1 ) + 1 2 ( x 4 − x 2 + 1 ) + 1 2 x 2 x 6 + 1 d x = 1 2 ∫ x 4 − 1 x 6 + 1 d x + 1 2 ∫ x 4 − x 2 + 1 x 6 + 1 d x + 1 2 ∫ x 2 x 6 + 1 d x = 1 4 3 ln ∣ x 2 + 1 − 3 x x 2 + 1 + 3 x ∣ + 1 2 arctan x + 1 6 arctan x 3 + C \int {{{{x^4}} \over {{x^6} + 1}}dx} = \int {{{{1 \over 2}\left( {{x^4} - 1} \right) + {1 \over 2}\left( {{x^4} - {x^2} + 1} \right) + {1 \over 2}{x^2}} \over {{x^6} + 1}}dx} = {1 \over 2}\int {{{{x^4} - 1} \over {{x^6} + 1}}dx} + {1 \over 2}\int {{{{x^4} - {x^2} + 1} \over {{x^6} + 1}}} dx + {1 \over 2}\int {{{{x^2}} \over {{x^6} + 1}}} dx = {1 \over {4\sqrt 3 }}\ln \left| {{{{x^2} + 1 - \sqrt 3 x} \over {{x^2} + 1 + \sqrt 3 x}}} \right| + {1 \over 2}\arctan x + {1 \over 6}\arctan {x^3} + C ∫x6+1x4dx=∫x6+121(x4−1)+21(x4−x2+1)+21x2dx=21∫x6+1x4−1dx+21∫x6+1x4−x2+1dx+21∫x6+1x2dx=431ln x2+1+3xx2+1−3x +21arctanx+61arctanx3+C
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