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Question Number 212046 by Spillover last updated on 28/Sep/24

Answered by Spillover last updated on 28/Sep/24

Answered by Spillover last updated on 28/Sep/24

Answered by Ghisom last updated on 28/Sep/24

$$\int (x^6+x^4+x^2)\sqrt{2x^4+3x^2+6}dx=$$$$={(2x^4+3x^2+6)}^{3/2}y$$$$$$$$\frac{d}{dx}\left[{(2x^4+3x^2+6)}^{3/2}y\right]=$$$$=\sqrt{2x^4+3x^2+6}((2x^4+3x^2+6)y^{\prime }+3x(4x^2+3)y)$$$$(2x^4+3x^2+6)y^{\prime }+3x(4x^2+3)y=x^6+x^4+x^2$$$$\Rightarrow y={ax}^3+{bx}^2+cx+d$$$$\Rightarrow$$$$(18a-1)x^6+$$$$+16{bx}^5+$$$$+(18a+14c-1)x^4+$$$$+(15b+12d)x^3+$$$$+(18a+12c-1)x^2+$$$$+(12b+9d)x+$$$$+6c=0$$$$\Rightarrow a=\frac{1}{18}b=c=d=0$$$$y=\frac{x^3}{18}$$$$$$$$\int (x^6+x^4+x^2)\sqrt{2x^4+3x^2+6}dx=$$$$=\frac{x^3{(2x^4+3x^2+6)}^{3/2}}{18}+C$$

Commented by Spillover last updated on 28/Sep/24

$$great$$

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