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I'm trying to solve this indefinite integral: $$\int \ln(2x+1)dx$$

I tried integration by parts, letting

$u=\ln(2x+1) \qquad dv=dx \qquad du= \frac 2 {2x+1}dx \qquad v=x$

Plugging these into the standard formula, I got

$$\int \ln(2x+1)dx \quad = \quad x\ln(2x+1)- \int \frac {2x} {2x+1} dx$$

Now the problem I ran into: I have no idea how to solve $\int \frac {2x} {2x+1} dx$. I tried to use integration by parts on this, but it produced another integral I had no idea how to solve (it was $\int \frac {2x^2} {2x+1} dx$, in case you're curious).

Can someone help me out? (The only two integration techniques I know for now are substitution and integration by parts)

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  • $\begingroup$ Perform long division behind the integral and it comes out ok, you will get $1$ and a remainder divided by $2x+1$. Easy to integrate $\endgroup$
    – imranfat
    Commented Mar 3, 2015 at 0:11

4 Answers 4

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Hint. Just write $$ \frac {2x} {2x+1}=\frac {2x+1-1} {2x+1}=1-\frac {1} {2x+1} $$ and integrate each term.

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  • $\begingroup$ Both your and abel's answers were very helpful, but I slightly preferred this because of its immediate clarity and simplicity. $\endgroup$
    – Asker
    Commented Mar 3, 2015 at 0:37
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you can integrate $$\int \frac{2x}{2x+1} \, dx = \int \left(1 - \frac1{2x+1}\right) \, dx = x - \frac 12 \ln|2x+1| + C$$

p.s. i think it would have been easier had you made the substitution $u = 2x+1$ at the very beginning. this trick works on integrals involving composition with $ax + b.$ here are some examples $\int \dfrac{1}{(ax+b)^2} \, dx, \int \sin (ax + b) \, dx, \int \ln (ax + b)\, dx, etc.$

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$$I=\int\ln(2x+1)dx$$ let $u=2x+1$. Therefore $du=2dx$, then $dx=du/2$. Which gives $$I=\frac1{2}\int\ln udu$$ $$I=\frac{u}{2}(\ln u-1)$$ $$I=\frac{2x+1}{2}(\ln(2x+1)-1)+C$$ And badda boom badda bing there's your integral.

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By parts,

$$\int\log(2x+1)\,dx=\left(x+\frac12\right)\log(2x+1)-2\int\frac{x+\dfrac12}{2x+1}dx=\left(x+\frac12\right)\log(2x+1)-x.$$

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