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Question Number 186263 by mokys last updated on 02/Feb/23

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

Answered by cortano1 last updated on 03/Feb/23

$$\:{let}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\:=\:\mathrm{tan}\:{t}\:\Rightarrow\mathrm{2}\sqrt{{x}}\:=\:\mathrm{cot}\:{t} \\$$$$\Rightarrow\mathrm{4}{x}\:=\:\mathrm{cot}\:^{\mathrm{2}} {t}\: \\$$$$\Rightarrow\mathrm{4}{dx}\:=\:\mathrm{2cot}\:{t}\:\left(−\mathrm{csc}^{\mathrm{2}} \:{t}\right)\:{dt} \\$$$${I}=\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {t}}}\:\left(−\mathrm{2cot}\:{t}\:\mathrm{csc}^{\mathrm{2}} {t}\right){dt} \\$$$${I}=\:−\mathrm{2}\int\:\mathrm{cos}\:{t}\:\frac{\mathrm{cos}\:{t}}{\mathrm{sin}\:{t}}\:.\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} {t}}\:{dt}\: \\$$$${I}=−\mathrm{2}\int\:\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {t}}{\mathrm{sin}\:^{\mathrm{3}} {t}}\:{dt} \\$$$${I}=−\mathrm{2}\int\left(\mathrm{csc}^{\mathrm{3}} {t}−\mathrm{csc}\:{t}\right)\:{dt}\: \\$$$$\\$$

Answered by Mathspace last updated on 03/Feb/23

$${let}\:\frac{\mathrm{1}}{\mathrm{4}{x}}={t}\:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{4}{t}}\:\Rightarrow \\$$$${I}=−\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{+\infty} \sqrt{\mathrm{1}+{t}}\left(\frac{−{dt}}{\mathrm{4}{t}^{\mathrm{2}} }\right) \\$$$$\mathrm{4}{I}=\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{+\infty} \frac{\sqrt{\mathrm{1}+{t}}}{{t}^{\mathrm{2}} }{dt}\:\:\:\left(\mathrm{1}+{t}={z}^{\mathrm{2}} \right) \\$$$$=\int_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\infty} \frac{{z}}{\left({z}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\left(\mathrm{2}{z}\right){dz} \\$$$$=\int_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} \:\:{z}\left(\frac{\mathrm{2}{z}}{\left({z}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\right){dz}\:\:\left({by}\:{parts}\right) \\$$$$=\left[−\frac{{z}}{{z}^{\mathrm{2}} −\mathrm{1}}\right]_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\infty} −\int_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\infty} \left(−\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{1}}\right){dz} \\$$$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\infty} \left(\frac{\mathrm{1}}{{z}−\mathrm{1}}−\frac{\mathrm{1}}{{z}+\mathrm{1}}\right){dz} \\$$$$=\mathrm{2}\sqrt{\mathrm{5}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\mid\right]_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\infty} \\$$$$=\mathrm{2}\sqrt{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{1}}{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}+\mathrm{1}}\mid \\$$$$=\mathrm{2}\sqrt{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{2}}\right) \\$$