How do I find the integral $\int x \cdot 2^{x} d x$ $intx*2^xdx$ ?

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How do I find the integral $\int x \cdot 2^{x} d x$ $intx*2^xdx$ ?

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Answer 1 · 2014-08-03T22:18:41

$\int x \cdot 2^{x} d x = \frac{x \cdot 2^{x}}{ln 2} - \frac{2^{x}}{{(ln 2)}^{2}} + C$ $int x * 2^x dx = (x*2^x)/ln2 - 2^x / (ln2)^2 + C$

Process:

This problem will require multiple $u$ $u$ -substitutions and an application of integration by parts.

First, it's important to know how to integrate exponentials with bases other than $e$ $e$ . The key is to convert them to base $e$ $e$ using laws of exponents.

Here is what I mean:

First, look at the $2^{x}$ $2^x$ inside our integral. We know that a logarithm is essentially the exponent required to raise a base to a certain number, so, if we want to rewrite $2$ $2$ as an exponential with base $e$ $e$ , all we need to do is raise $e$ $e$ to the natural log of $2$ $2$ .

In other words,

$2^{x} = {(e^{ln 2})}^{x}$ $2^x = (e^ln2)^x$ .

Using one of the laws of exponents, this is actually the same statement as:

$2^{x} = (e^{x ln 2})$ $2^x = (e^(xln2))$ .

Using this information, we can rewrite our integral as:

$\int x \cdot 2^{x} d x = \int x \cdot e^{x ln 2} d x$ $int x * 2^x dx = int x * e^(xln2) dx$

Now we will perform an integration by parts. Remember the formula:

$\int u d v = u v - \int v d u$ $int u dv = uv - int v du$

We will let $u = x$ $u = x$ and $d v = e^{x ln 2} d x$ $dv = e^(xln2)dx$ . Any time you have an integrand with something multiplied by $x$ $x$ to the $1$ $1$ st power, it's a good hint to pick $x$ $x$ for $u$ $u$ .

Therefore, $d u = d x$ $du = dx$ . To find $v$ $v$ , we will need to do a $u$ $u$ -substitution. Since we already are using the variable $u$ $u$ in the integration by parts formula, we will use the letter $q$ $q$ instead.

$v = \int e^{x ln 2} d x$ $v = int e^(xln2)dx$
Let $q = x ln 2$ $q = xln 2$

Therefore,

$v = \frac{1}{ln 2} \int ln 2 \cdot e^{x ln 2} d x$ $v = 1/ln2 int ln 2 * e^(x ln 2) dx$
$v = \frac{1}{ln 2} \int e^{q} d q$ $v = 1/ln2 int e^(q) dq$

The integral of $e^{x}$ $e^x$ is simply $e^{x}$ $e^x$ . So,

$v = \frac{1}{ln 2} e^{q}$ $v = 1/ln2 e^(q)$
$v = \frac{1}{ln 2} e^{x ln 2}$ $v = 1/ln2 e^(x ln 2)$

Now we can start plugging things into our integration by parts formula:

$\int x \cdot e^{x ln 2} d x = u v - \int v d u$ $int x*e^(xln2)dx = uv - int v du$
$\int x \cdot e^{x ln 2} d x = \frac{x \cdot e^{x ln 2}}{ln 2} - \int \frac{1}{ln 2} e^{x ln 2} d x$ $int x*e^(xln2)dx = (x*e^(x ln 2))/ln2 - int 1/ln2 e^(x ln 2) dx$

We will bring the $\frac{1}{ln 2}$ $1/ln2$ out of the integral, as it is a constant:

$\int x \cdot e^{x ln 2} d x = \frac{x \cdot e^{x ln 2}}{ln 2} - \frac{1}{ln 2} \int e^{x ln 2} d x$ $int x*e^(xln2)dx = (x*e^(x ln 2))/ln2 - 1/ln2 int e^(x ln 2) dx$

Evaluating this second integral should be a breeze, since we immediately notice that it is equal to $v$ $v$ . Don't forget the constant of integration:

$\int x \cdot e^{x ln 2} d x = \frac{x \cdot e^{x ln 2}}{ln 2} - \frac{e^{x ln 2}}{{(ln 2)}^{2}} + C$ $int x*e^(xln2)dx = (x*e^(x ln 2))/ln2 - (e^(x ln 2))/(ln2)^2 + C$

Now, it could be a good idea to rewrite all the exponentials with base $2$ $2$ instead of base $e$ $e$ :

$\int x \cdot 2^{x} d x = \frac{x \cdot 2^{x}}{ln 2} - \frac{2^{x}}{{(ln 2)}^{2}} + C$ $int x * 2^x dx = (x*2^x)/ln2 - 2^x / (ln2)^2 + C$

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