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How to integrate this: ln(2x+1) dx. I tried use integration by parts. But unsuccessful.
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ln(u) integrates to u ln(u) - u by tables :)
integration by parts does work it out; you just have to use 1 dx your "dv"
\[\int ln(u)du=u~lnu-\int \frac{1}{u}u~du\]
But, I go to this integral: X/(2x+1)dx and I not know how to integrate this ^^
\[\int \frac{2x}{2x+1} \text{d}x\]?
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Yes
\[ \frac{2x}{2x+1}=\frac{2x+1-1}{2x+1}=1-\frac{1}{2x+1} \]does that help?
Thanks! I had forgot to use this
np :)
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