Question

How do I evaluate ∫ ln(2x+1) dx?

Answer 1

Steps followed :

1) Think about how you can effectively remove the function like natural logs because you yet don't know the integration of these functions.

So, the best way is to multiply and divide by (2x+1) And then substitute log base e (2x+1) = t or any other variable and then proceed.

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Then, I suppose you can easily solve any question in integration with these thoughts and mindset

Answer 2

Answer:

Starting from your second to last line (your integration was fine, minus a few dxdx's in you integrals):

=xln(2x+1)+ln|(2x+1)12|−x+C(1)

(1)=xln⁡(2x+1)+ln⁡|(2x+1)12|−x+C

Good, up to this point... ↑↑.

So the error was in your last equality at the very end:

You made an error by ignoring the fact that the first term with ln(2x+1)ln⁡(2x+1) as a factor also has xx as a factor, so we cannot multiply the arguments of lnln to get ln(2x+1)3/2ln⁡(2x+1)3/2. What you could have done was first express xln(2x+1)=ln(2x+1)xxln⁡(2x+1)=ln⁡(2x+1)x and then proceed as you did in your answer, but your result will then agree with your text's solution.

Alternatively, we can factor out like terms.

=xln(2x+1)+12ln(2x+1)−x+C(1)

(1)=xln⁡(2x+1)+12ln⁡(2x+1)−x+C

=12⋅2xln(2x+1)+12ln(2x+1)⋅1−x+C

=12⋅2xln⁡(2x+1)+12ln⁡(2x+1)⋅1−x+C

Factoring out 12ln(2x+1)12ln⁡(2x+1) gives us

=(12ln(2x+1))⋅(2x+1)−x+C

=(12ln⁡(2x+1))⋅(2x+1)−x+C

=12(2x+1)ln(2x+1)−x+C