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Question Number 201229 by Calculusboy last updated on 02/Dec/23

∫ (1/( (√(1+Inx))))dx

$$\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\boldsymbol{{Inx}}}}\boldsymbol{{dx}} \\ $$

Commented by mnjuly1970 last updated on 02/Dec/23

In(x) or Ln(x)=?

$${In}\left({x}\right)\:{or}\:\:\mathrm{L}{n}\left({x}\right)=? \\ $$

Commented by Calculusboy last updated on 02/Dec/23

the first one

$$\boldsymbol{{the}}\:\boldsymbol{{first}}\:\boldsymbol{{one}} \\ $$

Answered by witcher3 last updated on 02/Dec/23

ln(x)=t =∫(e^t /( (√(1+t)))) =∫(e^(((√((1+t))))^2 −1) /1).2d((√(1+t)));(√(1+t))=x_1 =2∫e^(x_1 ^2 −1) dx =(2/e)∫e^x_1 ^2 dx_1 ;x_1 =it=(2/( (√e)))i∫e^(−t^2 ) dt =(√(π/e)).i erfc(t) (2/( (√π)))∫_0 ^x e^(−t^2 ) dt=erfc(x) =i(√(π/e)).erfc(−ix_1 )+c ∫(dx/( (√(1+ln(x)))))=c+i(√((π/e).))erfc(−i(√(1+ln(x))));c∈R

$$\mathrm{ln}\left(\mathrm{x}\right)=\mathrm{t} \\ $$$$=\int\frac{\mathrm{e}^{\mathrm{t}} }{\:\sqrt{\mathrm{1}+\mathrm{t}}} \\ $$$$=\int\frac{\mathrm{e}^{\left(\sqrt{\left(\mathrm{1}+\mathrm{t}\right)}\right)^{\mathrm{2}} −\mathrm{1}} }{\mathrm{1}}.\mathrm{2d}\left(\sqrt{\mathrm{1}+\mathrm{t}}\right);\sqrt{\mathrm{1}+\mathrm{t}}=\mathrm{x}_{\mathrm{1}} \\ $$$$=\mathrm{2}\int\mathrm{e}^{\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}} \mathrm{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{e}}\int\mathrm{e}^{\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} } \mathrm{dx}_{\mathrm{1}} ;\mathrm{x}_{\mathrm{1}} =\mathrm{it}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{e}}}\mathrm{i}\int\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt} \\ $$$$=\sqrt{\frac{\pi}{\mathrm{e}}}.\mathrm{i}\:\mathrm{erfc}\left(\mathrm{t}\right) \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}=\mathrm{erfc}\left(\mathrm{x}\right) \\ $$$$=\mathrm{i}\sqrt{\frac{\pi}{\mathrm{e}}}.\mathrm{erfc}\left(−\mathrm{ix}_{\mathrm{1}} \right)+\mathrm{c} \\ $$$$\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{ln}\left(\mathrm{x}\right)}}=\mathrm{c}+\mathrm{i}\sqrt{\frac{\pi}{\mathrm{e}}.}\mathrm{erfc}\left(−\mathrm{i}\sqrt{\mathrm{1}+\mathrm{ln}\left(\mathrm{x}\right)}\right);\mathrm{c}\in\mathbb{R} \\ $$$$ \\ $$

Commented by Calculusboy last updated on 02/Dec/23

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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