Chapter 6, Problem 1RE

To determine

To calculate: The infinite integral of ${\displaystyle \int \left(x+1\right)e^{x}dx}$ by use the method of integration by parts.

Answer to Problem 1RE

Solution:

The infinite integral of ${\displaystyle \int \left(x+1\right)e^{x}dx}$ is $x{e}^x+C$.

Explanation of Solution

Given Information:

The provided integral is ${\displaystyle \int \left(x+1\right)e^{x}dx}$.

Formula used:

The method of integration by parts:

If $v$ and $u$ are two differentiable function of $x$,

Then,

${\displaystyle \int udv}=uv-{\displaystyle \int vdu}$

Steps to solve the integral problems:

Step1: At first find the most complicated portion of the integrand and try to letter it as $dv$ so that it can fit a fundamental integration rule. Then, the remaining factor or factors of the integrand will be $u$.

Step2: First find the factor whose derivative is simple and consider it as $u$ and then the remaining factor or factors of the integrand will be $dv$ and $dv$ should always include the term $dx$ of the original integrand.

Calculation:

Recall the provided integral.

${\displaystyle \int \left(x+1\right)e^{x}dx}$

Observe from the above integrand that the simplest portion of the integrand is $x+1$.

Now consider $u=x+1$ and the remaining factors as $dv=e^{x}dx$.

Therefore,

$du=dx$

And,

$v=e^x$

Apply the integration by parts method.

${\displaystyle \int udv}=uv-{\displaystyle \int vdu}$

Substitute $x+1$ for $u$, $e^x$ for $v$, $dx$ for $du$ and solve.

$\begin{array}{c}{\displaystyle \int \left(x+1\right)e^{x}dx}=\left(x+1\right)e^x-{\displaystyle \int e^{x}dx}\\ =\left(x+1\right)e^x-e^x+C\\ =x{e}^x+e^x-e^x+C\\ =x{e}^x+C\end{array}$

Hence, the infinite integral of ${\displaystyle \int \left(x+1\right)e^{x}dx}$ is $x{e}^x+C$.