Question

$\large\red{\sf{Question}}$
Solve the following
${\bf{ {\huge \int} {e}^{ \sqrt{x} }dx }}$
​

Answer 1

Step-by-step explanation:

I=∫exdx

Put x=t

21x−1/2dx=dt

⇒dx=2tdt

So, I=2∫tetdt

=2[tet−∫etdt]

=2tet−2et+C

=2ex(x−1)+c

Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\rm {e}^{ \sqrt{x} } \: dx$

To evaluate this integral, we use method of Substitution.

So, substitute

$\rm \: \sqrt{x} = y$

$\rm \: x = {y}^{2}$

$\rm \: dx = 2y \: dy$

So, on substituting these values in above integral, we get

$\rm \: = \: \displaystyle\int\rm {e}^{y} \: 2y \: dy$

$\rm \: = \: 2\displaystyle\int\rm y{e}^{y} \: dy$

Now, on integrating by parts, we get

$\rm \: = 2\bigg(y\displaystyle\int\rm {e}^{y}dy - \displaystyle\int\rm \bigg[\dfrac{d}{dy}y\displaystyle\int\rm {e}^{y}dy \bigg]dy\bigg)$

$\rm \: = 2\bigg(y {e}^{y} - \displaystyle\int\rm \bigg[1 \times {e}^{y}\bigg]dy\bigg)$

$\rm \: = 2\bigg(y {e}^{y} - \displaystyle\int\rm {e}^{y}dy\bigg)$

$\rm \: = 2\bigg(y {e}^{y} - {e}^{y}\bigg) + c$

$\rm \: = \: 2{e}^{y}(y - 1) + c$

$\rm \: = \: 2{e}^{ \sqrt{x} }( \sqrt{x} - 1) + c$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\displaystyle\int\rm {e}^{ \sqrt{x} } dx = \: 2{e}^{ \sqrt{x} }( \sqrt{x} - 1) + c \: }}$

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Basic Concept Used :-

Integration by Parts

$\boxed{ \rm{ \:\displaystyle\int\rm (u.v)dx \: = \:u\displaystyle\int\rm vdx - \displaystyle\int\rm \bigg[\dfrac{d}{dx}u\displaystyle\int\rm vdx \bigg] dx}}$

where,

• u is the function u(x)

• v is the function v(x)

• u' is the derivative of the function u(x)

For integration by parts , u and v are chosen according to the word ILATE, where alphabets have the following meaning :

• I - Inverse trigonometric functions

• L - Logarithmic functions

• A - Arithmetic functions

• T - Trigonometric functions

• E- Exponential functions

The alphabet which comes first is taken as u and other as v.

Formulae Used :-

$\boxed{ \rm{ \:\displaystyle\int\rm {e}^{x}dx \: = \: {e}^{x} + c \: \: }}$

$\boxed{ \rm{ \: \frac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: }}$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$