# Chain rule method doesn't result with same answer as u-sub. Why?

$\int \ln \left(2x\right)\,\mathrm{d}x$ is the integral in question.

I know how to solve it with the chain rule. $\frac{1}{2x}\times 2x = \frac{1}{x}$

But, because I know $u$-sub method, I wish to use it upon this integral.

$\int \ln \left(2x\right)\,\mathrm{d}x$

$u = 2x$

$du = 2dx \Rightarrow (1/2)du = dx$

Thus tranforming the integral into: $(1/2)\int \ln u \, \mathrm{d}u$

Which gives us $\frac{1}{2u}$ and undoing the sub results in $\frac{1}{4x}$

Now, this can not be, thus I am at fault and I do not see how. Please, help explain where I am wrong.

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$\int\ln udu\ne\frac1u$. It's the other way round: $\frac1u$ is the derivative of $\ln u$, not the integral. – bof Oct 30 '13 at 5:52

We usually integrate something like $\int \ln(2x) \, dx$ by parts, as it is quite efficient.

We use the formula $\int u \,dv=uv-\int v\,du$.

Let

\begin{align*} u &=\ln(2x) \quad\quad v=x \\ du &=\frac{1}{x}\, dx \,\,\,\,\,\,\,\,\,\, dv=dx \end{align*}

Then \begin{align*} &uv-\int v\,du \\ =&x\ln(2x)-\int 1 \, dx \\ =&x\ln(2x)-x+c. \end{align*}

If you were to put yourself in a situation where, after a $u$ substitution, all you had to integrate was $\ln(u)$, you would probably integrate that by parts as well (and then back substitute to the same result if done correctly), thus diminishing your efforts to demonstrate an alternate method.

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$$\frac{1}{2}\int \ln(2x) d2x = \frac{1}{2} \ln(2x)2x - \frac{1}{2}\int2xd\ln(2x) = \frac{\ln(2x)2x}{2} - \int \frac{2x}{x}dx = \frac{\ln(2x)2x}{2} - 2x + C$$

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