Question Number 201229 by Calculusboy last updated on 02/Dec/23
$$\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\boldsymbol{{Inx}}}}\boldsymbol{{dx}} \\ $$
Commented by mnjuly1970 last updated on 02/Dec/23
$${In}\left({x}\right)\:{or}\:\:\mathrm{L}{n}\left({x}\right)=? \\ $$
Commented by Calculusboy last updated on 02/Dec/23
$$\boldsymbol{{the}}\:\boldsymbol{{first}}\:\boldsymbol{{one}} \\ $$
Answered by witcher3 last updated on 02/Dec/23
$$\mathrm{ln}\left(\mathrm{x}\right)=\mathrm{t} \\ $$$$=\int\frac{\mathrm{e}^{\mathrm{t}} }{\:\sqrt{\mathrm{1}+\mathrm{t}}} \\ $$$$=\int\frac{\mathrm{e}^{\left(\sqrt{\left(\mathrm{1}+\mathrm{t}\right)}\right)^{\mathrm{2}} −\mathrm{1}} }{\mathrm{1}}.\mathrm{2d}\left(\sqrt{\mathrm{1}+\mathrm{t}}\right);\sqrt{\mathrm{1}+\mathrm{t}}=\mathrm{x}_{\mathrm{1}} \\ $$$$=\mathrm{2}\int\mathrm{e}^{\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}} \mathrm{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{e}}\int\mathrm{e}^{\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} } \mathrm{dx}_{\mathrm{1}} ;\mathrm{x}_{\mathrm{1}} =\mathrm{it}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{e}}}\mathrm{i}\int\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt} \\ $$$$=\sqrt{\frac{\pi}{\mathrm{e}}}.\mathrm{i}\:\mathrm{erfc}\left(\mathrm{t}\right) \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}=\mathrm{erfc}\left(\mathrm{x}\right) \\ $$$$=\mathrm{i}\sqrt{\frac{\pi}{\mathrm{e}}}.\mathrm{erfc}\left(−\mathrm{ix}_{\mathrm{1}} \right)+\mathrm{c} \\ $$$$\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{ln}\left(\mathrm{x}\right)}}=\mathrm{c}+\mathrm{i}\sqrt{\frac{\pi}{\mathrm{e}}.}\mathrm{erfc}\left(−\mathrm{i}\sqrt{\mathrm{1}+\mathrm{ln}\left(\mathrm{x}\right)}\right);\mathrm{c}\in\mathbb{R} \\ $$$$ \\ $$
Commented by Calculusboy last updated on 02/Dec/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$