The variable $x$ and natural exponential function $e^x$ formed an algebraic function by their product, and we have to evaluate the indefinite integration of the function $xe^x$ with respect to $x$ in calculus.

$\displaystyle \int{xe^x \,} dx$

In this integral problem, we can notice that

- The algebraic functions $x$ and $e^x$ are involved in multiplication.
- The power of the algebraic function $x$ can reduced by differentiation.
- The integration of the natural exponential function $e^x$ can be evaluated directly.

The indefinite integration of the given algebraic function can be evaluated only by the integration by parts method.

Take, $\displaystyle \int{xe^x \,} dx$ $\,=\,$ $\displaystyle \int{u}dv$

In this indefinite integration problem, we use the power reduction technique for solving this problem. So, we must take $u = x$ and $dv = e^x dx$

Now, we have to evaluate the differential element $du$ by differentiation and the variable $v$ by the integration.

$u = x$

$\implies$ $\dfrac{du}{dx} = \dfrac{dx}{dx}$

$\,\,\, \therefore \,\,\,\,\,\,$ $du = dx$

$dv = e^x dx$

Now, solve the differential equation by using the integration rule of natural exponential function.

$\implies$ $\displaystyle \int{\,}dv = \int{e^x \,}dx$

$\implies$ $\displaystyle \int{\,}dv = \int{e^x \,}dx$

$\implies$ $v+c = e^x+c$

$\,\,\, \therefore \,\,\,\,\,\,$ $v = e^x$

Now, substitute the values of the variables and differentials in the formula of the integration of parts for evaluating the indefinite integration of the given algebraic function mathematically.

$\displaystyle \int{u}dv$ $\,=\,$ $uv$ $-$ $\displaystyle \int{v}du$

$\implies$ $\displaystyle \int{xe^x \,}dx$ $\,=\,$ $x \times e^x$ $-$ $\displaystyle \int{e^x \,}dx$

$\implies$ $\displaystyle \int{xe^x \,}dx$ $\,=\,$ $xe^x$ $-$ $\displaystyle \int{e^x \,}dx$

$\implies$ $\displaystyle \int{xe^x \,}dx$ $\,=\,$ $xe^x-e^x+c$

$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \int{xe^x \,}dx$ $\,=\,$ $e^x(x-1)+c$

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