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Question Number 177242 by peter frank last updated on 02/Oct/22

∫ ((3x^(16) +5x^(14) )/((x^5 +x^2 +1)^4 ))dx

$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{\mathrm{3x}^{\mathrm{16}} +\mathrm{5x}^{\mathrm{14}} }{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }\mathrm{dx} \\ $$$$ \\ $$

Answered by som(math1967) last updated on 02/Oct/22

∫((3x^(16) +5x^(14) )/({x^5 (1+(1/x^3 )+(1/x^5 ))}^4 ))dx ∫((x^(14) (3x^2 +5))/(x^(20) (1+(1/x^3 )+(1/x^5 ))^4 ))dx ∫(((3/x^4 )+(5/x^6 ))/((1+(1/x^3 )+(1/x^5 ))^4 ))dx let (1+(1/x^3 )+(1/x^5 ))=t ⇒−((3/x^4 )+(5/x^6 ))dx=dt −∫(dt/t^4 ) =−(t^(−4+1) /(−4+1)) +C =(1/(3t^3 )) +C =(1/(3(1+(1/x^3 )+(1/x^5 ))^3 )) +C =(x^(15) /(3(x^5 +x^2 +1)^3 )) +C

$$\:\int\frac{\mathrm{3}{x}^{\mathrm{16}} +\mathrm{5}{x}^{\mathrm{14}} }{\left\{{x}^{\mathrm{5}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)\right\}^{\mathrm{4}} }{dx} \\ $$$$\int\frac{{x}^{\mathrm{14}} \left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}\right)}{{x}^{\mathrm{20}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{4}} }{dx} \\ $$$$\int\frac{\frac{\mathrm{3}}{{x}^{\mathrm{4}} }+\frac{\mathrm{5}}{{x}^{\mathrm{6}} }}{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{4}} }{dx} \\ $$$$\:{let}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)={t} \\ $$$$\:\Rightarrow−\left(\frac{\mathrm{3}}{{x}^{\mathrm{4}} }+\frac{\mathrm{5}}{{x}^{\mathrm{6}} }\right){dx}={dt} \\ $$$$\:−\int\frac{{dt}}{{t}^{\mathrm{4}} } \\ $$$$=−\frac{{t}^{−\mathrm{4}+\mathrm{1}} }{−\mathrm{4}+\mathrm{1}}\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{3}} }\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} }\:+{C} \\ $$$$=\frac{{x}^{\mathrm{15}} }{\mathrm{3}\left({x}^{\mathrm{5}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:+{C} \\ $$

Commented by Ar Brandon last updated on 02/Oct/22

wow!

Commented by peter frank last updated on 02/Oct/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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