Integrate by Parts: Solving (xe^(2x))/((1+2x)^2)

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In summary, the conversation was about integrating (xe^(2x))/((1+2x)^2) using substitution and integration by parts techniques. One person suggested substituting w=1+2x and another person suggested using the quotient rule. The final solution involved letting u=xe^(2x).
  • #1
annie122
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how do i integrate
(xe^(2x))/((1+2x)^2)??
do i substitute 1 + 2x = w?

but if i do, how do i proceed from there?
 
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  • #2
Re: integation by parts

I would definitely recommend $w=1+2x$ to begin. Linear substitutions like that cost you nothing, and could gain you quite a bit, as in this case. What is the resulting $w$ integral?
 
  • #3
Re: integation by parts

i got 1/4(e^(w-1)/w - e^(w-1)/w^2))
but this doesn't look any easier... :/
 
  • #4
Re: integation by parts

Here, let me type it out $\LaTeX$ style:
$$\frac{1}{4e} \int \frac{e^{w}(w-1)}{w^{2}} \, dw.$$
Is that what you have?
 
  • #5
Re: integation by parts

yes, and i notice one technique you used, which is to take out the e^(-1), but i still don't see how i should go..

thx for the help btw :)EDIT:
actually i found the answer;
i had to let u = xe^(2x)
 
Last edited:
  • #6
Re: integation by parts

Another way to solve this integral is to notice that
\begin{align*}
\frac{d}{dx} \frac{f(x)}{g(x)}&=\frac{g(x)\, f'(x)-f(x) \, g'(x)}{(g(x))^{2}} \\
\frac{f(x)}{g(x)}+C&=\int \frac{g(x)\, f'(x)-f(x) \, g'(x)}{(g(x))^{2}} \, dx.
\end{align*}
The quotient rule doesn't come in handy all that often, but when it does, it surprises you.
 

FAQ: Integrate by Parts: Solving (xe^(2x))/((1+2x)^2)

What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are functions and du and dv are their respective differentials.

How do you choose which function to use as u and which to use as dv?

When using integration by parts, it is important to choose u and dv in a way that simplifies the integral. Typically, u should be chosen as the function that becomes simpler after taking its derivative, and dv should be chosen as the more complicated function that can be easily integrated.

Can integration by parts be used for all types of integrals?

No, integration by parts is most effective for integrals involving products of polynomials, exponential functions, and trigonometric functions. It may not be useful for integrals with more complicated functions such as logarithms or inverse trigonometric functions.

How do you apply integration by parts to the given problem, (xe^(2x))/((1+2x)^2)?

To solve this integral using integration by parts, we can choose u = x and dv = (e^(2x))/((1+2x)^2). Then, du = 1 and v = -1/(2(1+2x)). Plugging these values into the integration by parts formula, we get ∫(xe^(2x))/((1+2x)^2) dx = -x/(2(1+2x)) + ∫1/(2(1+2x)) dx. This integral can then be solved using basic integration techniques.

Are there any other methods for solving integrals besides integration by parts?

Yes, there are several other methods for solving integrals, including substitution, trigonometric substitution, partial fractions, and using integration tables. Each method is useful for different types of integrals and it is important to choose the method that best fits the problem at hand.

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