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
![∫(x^6 +x^4 +x^2 )(√(2x^4 +3x^2 +6)) dx= =(2x^4 +3x^2 +6)^(3/2) y (d/dx)[(2x^4 +3x^2 +6)^(3/2) y]= =(√(2x^4 +3x^2 +6))((2x^4 +3x^2 +6)y′+3x(4x^2 +3)y) (2x^4 +3x^2 +6)y′+3x(4x^2 +3)y=x^6 +x^4 +x^2 ⇒ y=ax^3 +bx^2 +cx+d ⇒ (18a−1)x^6 + +16bx^5 + +(18a+14c−1)x^4 + +(15b+12d)x^3 + +(18a+12c−1)x^2 + +(12b+9d)x+ +6c=0 ⇒ a=(1/(18)) b=c=d=0 y=(x^3 /(18)) ∫(x^6 +x^4 +x^2 )(√(2x^4 +3x^2 +6)) dx= =((x^3 (2x^4 +3x^2 +6)^(3/2) )/(18))+C](https://www.tinkutara.com/question/Q212059.png)
$$\int\left({x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{\mathrm{2}} \right)\sqrt{\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}}\:{dx}= \\ $$$$=\left(\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}\right)^{\mathrm{3}/\mathrm{2}} {y} \\ $$$$ \\ $$$$\frac{{d}}{{dx}}\left[\left(\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}\right)^{\mathrm{3}/\mathrm{2}} {y}\right]= \\ $$$$=\sqrt{\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}}\left(\left(\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}\right){y}'+\mathrm{3}{x}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\right){y}\right) \\ $$$$\left(\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}\right){y}'+\mathrm{3}{x}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\right){y}={x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:{y}={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{18}{a}−\mathrm{1}\right){x}^{\mathrm{6}} + \\ $$$$+\mathrm{16}{bx}^{\mathrm{5}} + \\ $$$$+\left(\mathrm{18}{a}+\mathrm{14}{c}−\mathrm{1}\right){x}^{\mathrm{4}} + \\ $$$$+\left(\mathrm{15}{b}+\mathrm{12}{d}\right){x}^{\mathrm{3}} + \\ $$$$+\left(\mathrm{18}{a}+\mathrm{12}{c}−\mathrm{1}\right){x}^{\mathrm{2}} + \\ $$$$+\left(\mathrm{12}{b}+\mathrm{9}{d}\right){x}+ \\ $$$$+\mathrm{6}{c}=\mathrm{0} \\ $$$$\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{18}}\:\:\:\:\:{b}={c}={d}=\mathrm{0} \\ $$$${y}=\frac{{x}^{\mathrm{3}} }{\mathrm{18}} \\ $$$$ \\ $$$$\int\left({x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{\mathrm{2}} \right)\sqrt{\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}}\:{dx}= \\ $$$$=\frac{{x}^{\mathrm{3}} \left(\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{18}}+{C} \\ $$
Commented by Spillover last updated on 28/Sep/24
$${great} \\ $$