What is the integral of $(\ln(4-2x))^2$?

Question

$$\int \ln^2(4-2x)\ dx$$

I have tried many ways to do this, but I have yet to come to an answer.

Answer 1

First make the substitution $u = 4 - 2x$. This is just a linear change of variables.

Now use the formula $$\int \ln^2 x dx = x \bigg(\ln^2 x - 2 \ln x +2 \bigg) +C $$

This formula can be obtained by integration by parts with $u = \ln^2 x$ and $dv = dx$, if I recall correctly.

Answer 2

First, start with a simple u-substitution; it is a linear term so after the change of variables it will not complicate the integral: $u=4-2x$

$$\int \ln^2 u\:\mathrm{d}u(-\frac{1}{2})=-\frac{1}{2}\int \ln^2u \: \mathrm{d}u$$

Then, integrate by parts($\int u\:\mathrm{d}v=uv-\int v\:\mathrm{d}u$); it is easy to define the easy to derivate one as the $u$ function, or the easy to integrate as the $\mathrm{d}v$. Since it is easy for us to derivate $\ln^2\: x$.

Then, $\int \ln^2 x\: \mathrm{d}x$, with $u=\ln^2 x\:\:\:\mathrm{d}u=2(\ln\:x)(\ln\:x)'=\frac{2\ln\:x}{x}\ln\:\:\:\:v=x\:\:\:\:\mathrm{d}v=1$:

$$\int \ln^2 x\: dx = x\ln^2x-\int2\ln\:x\: \mathrm{d}x$$

And by parts again,$u=\ln\:x\:\:\:\mathrm{d}u=\frac{1}{x}\ln\:\:\:\:v=x\:\:\:\:\mathrm{d}v=1$

$$x\ln^2x-\int2\ln\:x\: \mathrm{d}x = x\ln^2x-2\int \ln\:x\:\mathrm{d}x$$

$$x\ln^2x-\int2\ln\:x\: \mathrm{d}x = x\ln^2x-2(x\ln\:x-\int x\:\frac{1}{x}\mathrm{d}x) $$

$$x\ln^2x-\int2\ln\:x\: \mathrm{d}x = x\ln^2x-2(x\ln\:x-\int \mathrm{d}x) $$

$$x\ln^2x-\int2\ln\:x\: \mathrm{d}x = x\ln^2x-2(x\ln\:x-x+C) $$

Putting into the integral above,

$$-\frac{1}{2}\int \ln^2u \:\mathrm{d}u= -\frac{1}{2}\left [u\ln^2u-2(u\ln\:u-u+C) \right ]$$

Substitute back the $u$ and you are done.

Answer 3

Substitute $u = 2x - 4 \longrightarrow \text{d}x = \frac{1}{2}\text{d}u$ $$\int\ln^2({4 - 2x})~\text{d}x = \frac{1}{2}\int(\ln(u) + \ln(-1))^2~\text{d}u\tag{1}$$ Lets solve the inner part of the integral in $(1)$. Substitute $v = \ln(u) \longrightarrow \text{d}u = u\text{d}v$ $$\int(\ln(u) + \ln(-1))^2~\text{d}u = \int(v + \ln(-1))^2e^v~\text{d}u\tag{2}$$ Integrate by parts to get $$\int(v + \ln(-1))^2e^v~\text{d}u = (v + \ln(-1))^2e^v - 2\int (v + \ln(-1))^2e^v~\text{d}v\tag{3}$$ Lets solve the inner part of the integral in $(2)$. Integrate by parts to get $$\int (v + \ln(-1))^2e^v~\text{d}v = (v + \ln(-1))e^v - \int e^v~\text{d}v = (v + \ln(-1))e^v - e^v\tag{4}$$ Plug in $(4)$ in $(3)$, $$ \begin{align*}(v + \ln(-1))^2e^v - 2\int (v + \ln(-1))^2e^v~\text{d}v &= (v + \ln(-1))^2e^v - 2((v + \ln(-1))e^v - e^v)\\ &= (v + \ln(-1))^2e^v - 2(v + \ln(-1))e^v) + 2e^v\tag{5} \end{align*}$$ Reverse substitution in $(5)$, $$(v + \ln(-1))^2e^v - 2\int (v + \ln(-1))^2e^v~\text{d}v = u(\ln(u) + \ln(-1))^2 - 2u(\ln(u) + \ln(-1)) + 2u\tag{6}$$ Comparing $(3)$ and $(6)$,$$ \int(\ln(u) + \ln(-1))^2~\text{d}u = u(\ln(u) + \ln(-1))^2 - 2u(\ln(u) + \ln(-1)) + 2u\tag{7}$$ Plug in $(7)$ in (1), $$\frac{1}{2}\int(\ln(u) + \ln(-1))^2~\text{d}u = \frac{u(\ln(u) + \ln(-1))^2}{2} - u\ln(u) + \ln(-1) + u\tag{8}$$ Reverse substitution in $(8)$, $$\begin{align*}\frac{1}{2}&\int(\ln(u) + \ln(-1))^2~\text{d}u = \\&\frac{(2x - 4)(\ln(2x - 4) + \ln(-1))^2}{2} - (2x - 4)(\ln(2x - 4) + \ln(-1)) + 2x - 4\end{align*}$$ $$\begin{align*}\implies&\int\ln^2({4 - 2x})~\text{d}x = \\&\frac{(2x - 4)(\ln(2x - 4) + \ln(-1))^2}{2} - (2x - 4)(\ln(2x - 4) + \ln(-1)) + 2x - 4 + C\end{align*}$$