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Question Number 542 by 123456 last updated on 25/Jan/15

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\sqrt{{x}^{\mathrm{6}} +\mathrm{4}}}{dx} \\$$

Answered by prakash jain last updated on 24/Jan/15

$${x}^{\mathrm{3}} ={u} \\$$ $${x}^{\mathrm{2}} {dx}=\frac{{du}}{\mathrm{3}} \\$$ $${x}=\mathrm{0},\:{u}=\mathrm{0} \\$$ $${x}=\mathrm{1},\:{u}=\mathrm{1} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\sqrt{{x}^{\mathrm{6}} +\mathrm{4}}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\centerdot\frac{\mathrm{1}}{\sqrt{{u}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}{du} \\$$ $$=\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left({u}+\sqrt{{u}^{\mathrm{2}} +\mathrm{4}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$ $$=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)−\mathrm{ln}\:\mathrm{2}\right] \\$$ $$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\$$