Rinkoo Gupta
Last Activity: 10 Years ago
integral((x^2+1)+2)/x^6(1+x^2) .dx
=integraldx/x^6 +2integral dx/x^6(1+x^2)
=(x^(-5))/(-5) +2 integral dx/x^6(1+x^2)
put x =tan@ in second integral
=(-1/5).x^(-5) +2integral sec^2@/tan^6@(1+tan^2@) d@
=(-1/5)x^(-5)+2integral cot^6@d@
By using reduction formula for integral of cot^6@, we get
=(-1/5)x^(-5)+2[(-cot^5@)/5-integral cot^4@d@]
=(-1/5)x^(-5)+2[-1/5cot^5@-{-1/3cot^3@-integralcot^2@d@}]
=(-1/5)x^(-5)+2[-1/5cot^5@+1/3cot^3@+integral(cosec^2@-1)d@]
=(-1/5)x^(-5)+2[-1/5cot^5@+1/3cot^3@-cot@-@]] +C
=(-1/5)x^(-5)+2[-1/(5x^5) +1/(3.x^3) -1/x -arctanx] +C Ans.
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Rinkoo Gupta
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