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

Answered by qaz last updated on 09/Apr/22

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

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

$$\mathrm{thanks}$$

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