求不定积分:1、∫ dx/x^4(2x^2-1) 下面都是分母 2、∫x^4/(x+1)^100 dx
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求不定积分:1、∫ dx/x^4(2x^2-1) 下面都是分母 2、∫x^4/(x+1)^100 dx
第一题:
原式=-∫[1/x^2(2x^2-1)]d(1/x).
令1/x=u,则x=1/u.
∴原式=-∫{1/(1/u)^2[2(1/u)^2-1]}du
=-∫[u^4/(2-u^2)]du
=∫[(u^4-4+4)/(u^2-2)]du
=∫[(u^2+2)(u^2-2)/(u^2-2)]du+4∫[1/(u^2-2)]du
=∫(u^2+2)du+4∫[1/(u+√2)(u-√2)]du
=∫u^2du+2∫du+√2∫[(u+√2-u+√2)/(u+√2)(u-√2)]du
=(1/3)u^3+2u+√2∫[1/(u-√2)]du-√2∫[1/(u+√2)]du
=(1/3)(1/x)^3+2/x+√2ln|u-√2|-√2ln|u+√2|+C
=(1/3)/x^3+2/x+√2ln|1/x-√2|-√2ln|1/x+√2|+C
=(1/3)/x^3+2/x+√2ln|1-√2x|-√2ln|1+√2x|+C
第二题:
令x+1=u,得:x=u-1,dx=du.
∴原式=∫[(u-1)^4/u^100]du
=∫[(u^2-2u+1)^2/u^100]du
=∫[(u^4+4u^2+1-4u^3-4u+2u^2)/u^100]du
=∫(1/u^96)du+6∫(1/u^98)du+∫(1/u^100)du-4∫(1/u^97)du-4∫(1/u^99)du
=-(1/95)/u^95-(6/97)/u^97-(1/99)/u^99+(4/96)/u^96+(4/98)/u^98+C
=-1/[95(x+1)^95]+1/[24(x+1)^96]-6/[97(x+1)^97]
+2/[49(x+1)^98]-1/[99(x+1)^99]+C
原式=-∫[1/x^2(2x^2-1)]d(1/x).
令1/x=u,则x=1/u.
∴原式=-∫{1/(1/u)^2[2(1/u)^2-1]}du
=-∫[u^4/(2-u^2)]du
=∫[(u^4-4+4)/(u^2-2)]du
=∫[(u^2+2)(u^2-2)/(u^2-2)]du+4∫[1/(u^2-2)]du
=∫(u^2+2)du+4∫[1/(u+√2)(u-√2)]du
=∫u^2du+2∫du+√2∫[(u+√2-u+√2)/(u+√2)(u-√2)]du
=(1/3)u^3+2u+√2∫[1/(u-√2)]du-√2∫[1/(u+√2)]du
=(1/3)(1/x)^3+2/x+√2ln|u-√2|-√2ln|u+√2|+C
=(1/3)/x^3+2/x+√2ln|1/x-√2|-√2ln|1/x+√2|+C
=(1/3)/x^3+2/x+√2ln|1-√2x|-√2ln|1+√2x|+C
第二题:
令x+1=u,得:x=u-1,dx=du.
∴原式=∫[(u-1)^4/u^100]du
=∫[(u^2-2u+1)^2/u^100]du
=∫[(u^4+4u^2+1-4u^3-4u+2u^2)/u^100]du
=∫(1/u^96)du+6∫(1/u^98)du+∫(1/u^100)du-4∫(1/u^97)du-4∫(1/u^99)du
=-(1/95)/u^95-(6/97)/u^97-(1/99)/u^99+(4/96)/u^96+(4/98)/u^98+C
=-1/[95(x+1)^95]+1/[24(x+1)^96]-6/[97(x+1)^97]
+2/[49(x+1)^98]-1/[99(x+1)^99]+C
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