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Hint: - Apply integration by parts method to reach the answer which is given as

$\int {\left( {{I_1}} \right)\left( {{I_2}} \right)dx = {I_1}} \int {\left( {{I_2}} \right)dx - \int {\left( {\left( {\dfrac{d}{{dx}}{I_1}} \right)\int {\left( {{I_2}} \right)dx} } \right)} } dx$

Given integral is

$I = \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} $

Now break the integration into two parts

$ \Rightarrow I = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx + \int {{e^x}\left\{ {f'\left( x \right)} \right\}dx} } $

Let $I = {I_1} + {I_2}................................\left( 1 \right)$

First solve ${I_1}$

${I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} $

Now integrate it by parts where $f\left( x \right)$ is the first part and ${e^x}$ is the second part.

$ \Rightarrow {I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} = f\left( x \right)\int {{e^x}dx - \int {\left( {\left( {\dfrac{d}{{dx}}f\left( x \right)} \right)\left( {\int {{e^x}dx} } \right)} \right)dx} } $

As we know$\int {{e^x}dx = {e^x},{\text{ }}} \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right)$, so apply this property

$ \Rightarrow {I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} = {e^x}f\left( x \right) - \int {{e^x}f'\left( x \right)dx + c} $, (where c is some arbitrary integration constant)

From equation (1)

$

I = {I_1} + {I_2} = {e^x}f\left( x \right) - \int {{e^x}f'\left( x \right)dx + c + } \int {{e^x}\left\{ {f'\left( x \right)} \right\}dx} \\

\Rightarrow I = {e^x}f\left( x \right) + c \\

$

Hence, option (b) is correct.

Note: - In such types of questions always remember the key rule that first breaks the integration into parts, then integrate first integration by integration by parts method and leave the second integration as it is, after that it cancels out the second integration, doing this we will get the required answer.

$\int {\left( {{I_1}} \right)\left( {{I_2}} \right)dx = {I_1}} \int {\left( {{I_2}} \right)dx - \int {\left( {\left( {\dfrac{d}{{dx}}{I_1}} \right)\int {\left( {{I_2}} \right)dx} } \right)} } dx$

Given integral is

$I = \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} $

Now break the integration into two parts

$ \Rightarrow I = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx + \int {{e^x}\left\{ {f'\left( x \right)} \right\}dx} } $

Let $I = {I_1} + {I_2}................................\left( 1 \right)$

First solve ${I_1}$

${I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} $

Now integrate it by parts where $f\left( x \right)$ is the first part and ${e^x}$ is the second part.

$ \Rightarrow {I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} = f\left( x \right)\int {{e^x}dx - \int {\left( {\left( {\dfrac{d}{{dx}}f\left( x \right)} \right)\left( {\int {{e^x}dx} } \right)} \right)dx} } $

As we know$\int {{e^x}dx = {e^x},{\text{ }}} \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right)$, so apply this property

$ \Rightarrow {I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} = {e^x}f\left( x \right) - \int {{e^x}f'\left( x \right)dx + c} $, (where c is some arbitrary integration constant)

From equation (1)

$

I = {I_1} + {I_2} = {e^x}f\left( x \right) - \int {{e^x}f'\left( x \right)dx + c + } \int {{e^x}\left\{ {f'\left( x \right)} \right\}dx} \\

\Rightarrow I = {e^x}f\left( x \right) + c \\

$

Hence, option (b) is correct.

Note: - In such types of questions always remember the key rule that first breaks the integration into parts, then integrate first integration by integration by parts method and leave the second integration as it is, after that it cancels out the second integration, doing this we will get the required answer.

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