Integrate by parts: ∫ln(2x+1)dx
This question has answers here and I was wondering about the last answer to the question which is the way I did it, got a different answer?
https://math.stackexchange.com/a/1598118/372659
Here is the answer,
Is doing u-substitution, and then integration by parts a wrong method here? Because in my textbook the answer there is a ones in the other answers.
Answer
You should find
[xln(2x+1)]−∫x22x+1dx
=xln(2x+1)−∫2x+1−12x+1dx
=xln(2x+1)−x+12ln(2x+1)+C
=2x+12ln(2x+1)−x+C
You can also put u=2x+1 to get
∫ln(u)du2=12(uln(u)−u)+K =2x+12ln(2x+1)−x−12+K
with C=K−1/2.
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