Integration By Parts: Need help with a step

In summary, the conversation discusses integration by parts and specifically evaluates the integral of ln(2x + 1). The question asker is stuck on a step where the solution manual shows the numerator simplifying to 2x, and they are unsure of how that simplification occurred. After clarification, it is determined that it is a simple arithmetic step that results in the same numerator.
  • #1
daviddee305
16
0
Integration By Parts: Need help with a step...

Evaluate the integral:

[tex]\int[/tex] ln(2x + 1)dx

I worked it out up until:

Xln(2x + 1) - [tex]\int[/tex] 2x/(2x + 1) dx

Then the next step throws me off. I attached a scan from the solutions manual and circled the part that confused me. Could somebody explain how we went from "2x/(2x + 1)" to "[(2x + 1) - 1]/(2x + 1)"?

I'm sure it's a simple arithmetic step I'm overlooking, which is why I'm reviewing a few weeks before Calc. 2.


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  • #2


Not much happens really, [itex](2x+1)-1=2x+1-1=2x[/itex]. As you can see both numerators are the same, the latter is just written in a more useful form.
 
  • #3


thanks... it makes sense now...
 

Related to Integration By Parts: Need help with a step

1. What is integration by parts?

Integration by parts is a method used in calculus to evaluate integrals of products of functions. It is based on the product rule of differentiation and allows us to rewrite an integral in a different form that may be easier to evaluate.

2. When should I use integration by parts?

Integration by parts is typically used when the integral contains a product of two functions that cannot be easily integrated using other methods (such as substitution or trigonometric identities). It is also useful when the integral contains a polynomial multiplied by a transcendental function (such as e^x or ln x).

3. How do I choose which function to differentiate and which to integrate?

When using integration by parts, you should choose the function to differentiate (u) based on the acronym "LIATE": logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. The function to integrate (dv) should then be the remaining function in the integral.

4. What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du. In words, it states that the integral of the product of two functions is equal to the first function multiplied by the antiderivative of the second function, minus the integral of the antiderivative of the first function multiplied by the second function.

5. How do I know when to stop using integration by parts?

You should continue using integration by parts until the integral on the right side of the formula becomes simpler or until you reach an integral that you know how to solve. This may require multiple iterations of the formula.

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