1-1-Inx-dx-

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

\int \frac{1}{\sqrt{1 + I n x}} d x

Commented by mnjuly1970 last updated on 02/Dec/23

I n (x) o r L n (x) = ?

Commented by Calculusboy last updated on 02/Dec/23

t h e f i r s t o n e

Answered by witcher3 last updated on 02/Dec/23

\ln (x) = t

= \int \frac{e^{t}}{\sqrt{1 + t}}

= \int \frac{e^{{(\sqrt{(1 + t)})}^{2} - 1}}{1} . 2 d (\sqrt{1 + t}); \sqrt{1 + t} = x_{1}

= 2 \int e^{x_{1}^{2} - 1} dx

= \frac{2}{e} \int e^{x_{1}^{2}} {dx}_{1}; x_{1} = it = \frac{2}{\sqrt{e}} i \int e^{- t^{2}} dt

= \sqrt{\frac{π}{e}} . i erfc (t)

\frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} dt = erfc (x)

= i \sqrt{\frac{π}{e}} . erfc (- {ix}_{1}) + c

\int \frac{dx}{\sqrt{1 + \ln (x)}} = c + i \sqrt{\frac{π}{e} .} erfc (- i \sqrt{1 + \ln (x)}); c \in R

Commented by Calculusboy last updated on 02/Dec/23

t h a n k s s i r