Question Number 72346 by aliesam last updated on 27/Oct/19
Commented by mathmax by abdo last updated on 27/Oct/19
![I =∫_0 ^∞ ((ln(x^(10) +x^6 +x^4 +1))/((x+1)^3 ))dx by psrts u^′ =(x+1)^(−3) and v= ln(x^(10) +x^6 +x^4 +1) ⇒I=[−(1/2)(x+1)^(−2) ln(x^(10) +x^6 +x^4 +1)]_0 ^(+∞) −∫_0 ^∞ (−(1/2))(x+1)^(−2) ×((10x^2 +6x^5 +4x^3 )/(x^(10) +x^6 +x^4 +1))dx =0 +(1/2)∫_0 ^∞ ((10x^2 +6x^5 +4x^3 )/((x+2)^2 (x^(10) +x^6 +x^4 +1)))dx rest to decompose F(x)=((10x^2 +6x^5 +4x^3 )/((x+2)^2 (x^(10) +x^6 +x^4 +1))) ...be continued...](Q72350.png)
$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({x}^{\mathrm{10}} +{x}^{\mathrm{6}} \:+{x}^{\mathrm{4}} \:+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }{dx}\:\:\:{by}\:{psrts}\:{u}^{'} =\left({x}+\mathrm{1}\right)^{−\mathrm{3}} \:{and}\:{v}= \\ $$$${ln}\left({x}^{\mathrm{10}} \:+{x}^{\mathrm{6}} \:+{x}^{\mathrm{4}} \:+\mathrm{1}\right)\:\Rightarrow{I}=\left[−\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}\right)^{−\mathrm{2}} {ln}\left({x}^{\mathrm{10}} \:+{x}^{\mathrm{6}} \:+{x}^{\mathrm{4}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{+\infty} \\ $$$$−\int_{\mathrm{0}} ^{\infty} \:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}+\mathrm{1}\right)^{−\mathrm{2}} ×\frac{\mathrm{10}{x}^{\mathrm{2}} \:+\mathrm{6}{x}^{\mathrm{5}} \:+\mathrm{4}{x}^{\mathrm{3}} }{{x}^{\mathrm{10}} \:+{x}^{\mathrm{6}} \:+{x}^{\mathrm{4}} \:+\mathrm{1}}{dx} \\ $$$$=\mathrm{0}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{10}{x}^{\mathrm{2}} \:+\mathrm{6}{x}^{\mathrm{5}} \:+\mathrm{4}{x}^{\mathrm{3}} }{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \left({x}^{\mathrm{10}} \:+{x}^{\mathrm{6}} \:+{x}^{\mathrm{4}} \:+\mathrm{1}\right)}{dx}\:{rest}\:{to}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{\mathrm{10}{x}^{\mathrm{2}} \:+\mathrm{6}{x}^{\mathrm{5}} \:+\mathrm{4}{x}^{\mathrm{3}} }{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \left({x}^{\mathrm{10}} \:+{x}^{\mathrm{6}} \:+{x}^{\mathrm{4}} \:+\mathrm{1}\right)}\:...{be}\:{continued}... \\ $$