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

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.

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

### Deciding the method of evaluating the integration

In this integral problem, we can notice that

1. The algebraic functions $x$ and $e^x$ are involved in multiplication.
2. The power of the algebraic function $x$ can reduced by differentiation.
3. 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, $\displaystyle \int{xe^x \,} dx$ $\,=\,$ $\displaystyle \int{u}dv$

### Prepare the elements for indefinite integration

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.

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

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

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

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

### Evaluate the integration by the integration by parts

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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