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Question Number 168368 by SUPERMATH last updated on 09/Apr/22

Answered by qaz last updated on 09/Apr/22

∫(x^2 /( (√(1+x+x^2 ))))dx=∫(√(1+x+x^2 ))−((1+x)/( (√(1+x+x^2 ))))dx  =x(√(1+x+x^2 ))−∫((x+2x^2 )/( 2(√(1+x+x^2 ))))dx−∫((1+x)/( (√(1+x+x^2 ))))dx  =x(√(1+x+x^2 ))−∫((x^2 +1+(3/2)x)/( (√(1+x+x^2 ))))dx  =(1/2)x(√(1+x+x^2 ))−(3/8)∫((1+2x)/( (√(1+x+x^2 ))))dx−(1/8)∫(dx/( (√(1+x+x^2 ))))  =(1/2)x(√(1+x+x^2 ))−(3/4)(√(1+x+x^2 ))−(1/8)ln(x+(1/2)+(√(1+x+x^2 )))+C

$$\int\frac{\mathrm{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\int\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}+\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$=\mathrm{x}\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\int\frac{\mathrm{x}+\mathrm{2x}^{\mathrm{2}} }{\:\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}−\int\frac{\mathrm{1}+\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$=\mathrm{x}\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\int\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{8}}\int\frac{\mathrm{1}+\mathrm{2x}}{\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{4}}\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }\right)+\mathrm{C} \\ $$

Commented by peter frank last updated on 09/Apr/22

thanks

$$\mathrm{thanks} \\ $$

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