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Question Number 72346 by aliesam last updated on 27/Oct/19

Commented by mathmax by abdo last updated on 27/Oct/19

$$I={\int }_0^{\mathrm{\infty }}\frac{ln(x^{10}+x^6+x^4+1)}{{(x+1)}^3}dxbypsrts{u}^{{}^{\prime }}={(x+1)}^{-3}andv=$$$$ln(x^{10}+x^6+x^4+1)\Rightarrow I={[-\frac{1}{2}{(x+1)}^{-2}ln(x^{10}+x^6+x^4+1)]}_0^{+\mathrm{\infty }}$$$$-{\int }_0^{\mathrm{\infty }}(-\frac{1}{2}){(x+1)}^{-2}\times \frac{10x^2+6x^5+4x^3}{x^{10}+x^6+x^4+1}dx$$$$=0+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{10x^2+6x^5+4x^3}{{(x+2)}^2(x^{10}+x^6+x^4+1)}dxresttodecompose$$$$F\left(x\right)=\frac{10x^2+6x^5+4x^3}{{(x+2)}^2(x^{10}+x^6+x^4+1)}...becontinued...$$

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