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Integrate by parts: $\int \ln (2x + 1) \, dx$

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.

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    $\begingroup$ Not sure why the current answers below did not read beyond the title of your question. Anyway, your approach is fine, and yields the same answer (the $-1/2$ can be absorbed into the $+C$); perhaps the original question was some sort of homework exercise to practice using integration by parts. $\endgroup$ – angryavian Feb 11 '18 at 17:06
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    $\begingroup$ You can write $u = 2x+1$ so that $du=2\,dx$ and $\dfrac 1 2 \, du = dx,$ and then you get $$ \int\ln(2x+1)\,dx = \frac 1 2 \int \ln u\,du, $$ but then you still need to integrate by parts. However, you can go straight into integration by parts, thus $$ \int \ln(2x+1)\,dx = \int u\, dx = xu - \int x\,du $$ $$ = x\ln(2x+1) - \int x \frac 2 {2x+1} \, dx $$ and so on. $\endgroup$ – Michael Hardy Feb 11 '18 at 17:06
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You should find

$$[x\ln (2x+1)]-\int x\frac {2}{2x+1}dx $$

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

$$=x\ln (2x+1)-x+\frac {1}{2}\ln (2x+1) +C$$

$$=\frac {2x+1}{2}\ln (2x+1)-x+C $$

You can also put $u=2x+1$ to get

$$\int \ln (u)\frac {du}{2}=\frac {1}{2}(u\ln (u)-u)+K $$ $$=\frac {2x+1}{2}\ln (2x+1)-x-\frac {1}{2}+K $$

with $C=K-1/2$.

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we have $$\int\ln(2x+1)dx=x\ln(2x+1)-\int\frac{2x}{2x+1}dx$$ to solve

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