Why $\int \frac{1}{2x+1}\mathrm dx\neq \ln |2x+1|$?

Question

Why $$\int \frac{1}{2x+1}\mathrm dx\neq \ln |2x+1|$$?

While doing integration by partial fraction. I noticed they wrote that $$\int \frac{1}{2x+1}\mathrm dx=\frac{1}{2} \int \frac{2}{2x+1}=\frac{1}{2}\ln |2x+1|$$

Why $2$ is needed in numerator for this type of integration? There's only single $2$ in denominator.

Answer 1

If $F'(x)=f(x)$, then, for any two numbers $a$ and $b$, if you differentiate $F(ax+b)$, what you get is $af(ax+b)$ (by the chain rule).

So, since, when you differentiate $\ln|x|$, you get $\frac1x$, when you differentiate $\ln|2x+1|$, you get $\frac2{2x+1}$. Therefore, if what you want to get is $\frac1{2x+1}$, then differentiate $\frac12\ln|2x+1|$.

Answer 2

$$u=2x+1, \mathrm{du}=2\times \mathrm{dx} , \frac{\mathrm{du}}{2}=\mathrm{dx}$$

$$\int \frac{1}{2x+1}\mathrm dx =\int \frac{\frac{\mathrm{du}}{2}}{u}=\frac{1}{2}\int \frac{\mathrm{du}}{u}=\frac{1}{2}\ln{|u|}+c=\frac{1}{2}\ln|2x+1|+c$$

Answer 3

I like to think of it as (in general): $$\int\frac{1}{ax+b}dx=\frac1a\int\frac{a}{ax+b}dx=\frac1a\ln|ax+b|+C$$ basically because we can use the rule that: $$\int\frac{f'(x)}{f(x)}dx=\ln|f|+C$$ which you can justify by making the substitution $u=f(x)$ or noticing that: $$\frac{f'}{f}dx=\frac{df}{dx}\frac{dx}{f}=\frac{df}{f}$$

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so since $(2x+1)'=2$ we must "put 2 on top" but then to make the expression the same divide by two, giving: $$\frac12\int\frac2{2x+1}dx$$

Answer 4

Another way of looking at it: $$\int \frac{1}{2x+1}\mathrm dx= \\ \frac12\int \frac2{2x+1}\mathrm d(x)=\\ \frac12\int \frac1{2x+1}\mathrm d(2x)=\\ \frac12\int \frac1{t+1}\mathrm dt=\\ \frac12\ln |t+1|+C=\\ \frac12\ln|2x+1|+C.$$