Question

$\huge \dag \sf{No \: spamming}$​

Answer 1

Answer:

${\boxed{\boxed{\begin{array}{cc}\bf \: \to \:given : \\ \\ \displaystyle \int \rm \: \frac{ {e}^{ \sqrt{x} } }{ \sqrt{x} } \: \: dx \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: substitute \: \: u = \sqrt{x} \\ \\ \rm \implies \: \frac{du}{dx} = \frac{d}{dx}. \sqrt{x} = \frac{1}{2 \sqrt{x} } \\ \\ \therefore \rm \: dx = 2 \sqrt{x}. \: du \end{array}}}} \\ \\ \rm =\displaystyle \int \rm \: \frac{ {e}^{u} }{u} .2 \sqrt{x} \: \: du \\ \\ \rm = 2\displaystyle \int \rm \: \frac{ {e}^{u}}{ \cancel{u}} . \cancel{u }\: \: \: du \\ \\ \rm = 2\displaystyle \int \rm \: {e}^{u} \: du \\ \\ \rm \: = 2 {e}^{u} + c \\ \\ \orange{ \sf \: undo \: our \: substitution} \\ \\ \rm = 2 {e}^{ \sqrt{x} } + c \\ \\ \\ \blue{ \boxed{\therefore \displaystyle \int \rm \: \frac{ {e}^{ \sqrt{x} } }{ \sqrt{x} } \: dx = 2 {e}^{ \sqrt{x} } + c}} \\ \\ \sf \: c = integral \: constant\end{array}}}}$

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Answer 2

Step-by-step explanation:

Answer:

$\begin{gathered}{\boxed{\boxed{\begin{array}{cc}\bf \: \to \:given : \\ \\ \displaystyle \int \rm \: \frac{ {e}^{ \sqrt{x} } }{ \sqrt{x} } \: \: dx \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: substitute \: \: u = \sqrt{x} \\ \\ \rm \implies \: \frac{du}{dx} = \frac{d}{dx}. \sqrt{x} = \frac{1}{2 \sqrt{x} } \\ \\ \therefore \rm \: dx = 2 \sqrt{x}. \: du \end{array}}}} \\ \\ \rm =\displaystyle \int \rm \: \frac{ {e}^{u} }{u} .2 \sqrt{x} \: \: du \\ \\ \rm = 2\displaystyle \int \rm \: \frac{ {e}^{u}}{ \cancel{u}} . \cancel{u }\: \: \: du \\ \\ \rm = 2\displaystyle \int \rm \: {e}^{u} \: du \\ \\ \rm \: = 2 {e}^{u} + c \\ \\ \orange{ \sf \: undo \: our \: substitution} \\ \\ \rm = 2 {e}^{ \sqrt{x} } + c \\ \\ \\ \blue{ \boxed{\therefore \displaystyle \int \rm \: \frac{ {e}^{ \sqrt{x} } }{ \sqrt{x} } \: dx = 2 {e}^{ \sqrt{x} } + c}} \\ \\ \sf \: c = integral \: constant\end{array}}}}\end{gathered}$

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