Question-120773

Question Number 120773 by shaker last updated on 02/Nov/20

Answered by 675480065 last updated on 02/Nov/20

let u = \sqrt{x}

\Rightarrow du = \frac{1}{2 u} dx

\Rightarrow dx = 2 udu

\Rightarrow I = \int \frac{e^{u} . e^{u^{2}}}{u} . 2 udu

\Rightarrow I = 2 \int e^{u} . e^{u^{2}} du = 2 \int e^{u (1 + u)} du

w a i t \dots \dots .

Commented by 675480065 last updated on 02/Nov/20

= \frac{\sqrt{π} \erf (x + \frac{1}{2})}{{2 e}^{\frac{1}{4}}} + k