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Question Number 83921 by jagoll last updated on 08/Mar/20

∫  ((2x^(12) +5x^9 )/((x^5 +x^3 +1)^3 )) dx = ?

$$\int\:\:\frac{\mathrm{2x}^{\mathrm{12}} +\mathrm{5x}^{\mathrm{9}} }{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{dx}\:=\:? \\ $$

Answered by john santu last updated on 08/Mar/20

∫  ((2x^(12) +5x^9 )/((x^5 +x^3 +1)^3 )) ÷ (x^(15) /x^(15) ) dx =   ∫  (((2/x^3 )+(5/x^6 ))/((1+(1/x^2 )+(1/x^5 ))^3 )) dx  let u = 1+(1/x^2 )+(1/x^5 )  du = −(2/x^3 )−(5/x^6 ) dx   ⇔ ∫ ((−du)/u^3 ) = − ∫ u^(−3)  du  = (1/(2u^2 )) + c = (1/(2(1+(1/x^2 )+(1/x^5 ))^2 )) +c  =(x^(10) /(2(x^5 +x^3 +1)^2 )) + c

$$\int\:\:\frac{\mathrm{2x}^{\mathrm{12}} +\mathrm{5x}^{\mathrm{9}} }{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}} }\:\boldsymbol{\div}\:\frac{\mathrm{x}^{\mathrm{15}} }{\mathrm{x}^{\mathrm{15}} }\:\mathrm{dx}\:=\: \\ $$$$\int\:\:\frac{\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{6}} }}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }\right)^{\mathrm{3}} }\:\mathrm{dx} \\ $$$$\mathrm{let}\:\mathrm{u}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} } \\ $$$$\mathrm{du}\:=\:−\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{3}} }−\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{6}} }\:\mathrm{dx}\: \\ $$$$\Leftrightarrow\:\int\:\frac{−\mathrm{du}}{\mathrm{u}^{\mathrm{3}} }\:=\:−\:\int\:\mathrm{u}^{−\mathrm{3}} \:\mathrm{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2u}^{\mathrm{2}} }\:+\:\mathrm{c}\:=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }\right)^{\mathrm{2}} }\:+\mathrm{c} \\ $$$$=\frac{\mathrm{x}^{\mathrm{10}} }{\mathrm{2}\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }\:+\:\mathrm{c}\: \\ $$

Commented by jagoll last updated on 08/Mar/20

thank you

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

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