How do I find the closed form of this integral $\int_0^2\frac{\ln x}{x^3-2x+4}dx$?

Question

How do I find the closed form of this integral: $$I=\int_0^2\frac{\ln x}{x^3-2x+4}dx$$ First, I have a partial fraction of it: $$\frac{1}{x^3-2x+4}=\frac{1}{(x+2)(x^2-2x+2)}=\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+2}$$ $$A=\frac{1}{(x^3-2x+4)'}|_{x=-2}=\frac{1}{(3x^2-2)}|_{x=-2}=\frac{1}{10}$$ $$Bx+C=\frac{1}{x+2}|_{x^2-2x=-2}=\frac{1}{(x+2)}\frac{(x-4)}{(x-4)}|_{x^2-2x=-2}=$$ $$=\frac{(x-4)}{(x^2-2x-8)}|_{x^2-2x=-2}=\frac{(x-4)}{(-2-8)}|_{x^2-2x=-2}=-\frac{1}{10}(x-4)$$ Thus: $$\frac{1}{x^3-2x+4}=\frac{1}{10}\left(\frac{1}{x+2}-\frac{x-4}{x^2-2x+2}\right)$$ $$I=\frac{1}{10}\left(\int_0^2\frac{\ln x}{x+2}dx-\int_0^2\frac{(x-4)\ln x}{x^2-2x+2}dx\right)$$ What should I do next?

Answer 1

$\displaystyle I=\int_0^2\frac{\ln x}{x^3-2x+4}\,dx$

$\begin{align}I=\frac{1}{10}\int_0^2 \frac{\ln x}{2+x}\,dx+\frac{4}{10}\int_0^2 \frac{\ln x}{x^2-2x+2}\,dx-\frac{1}{10}\int_0^2 \frac{x\ln x}{x^2-2x+2}\,dx\end{align}$

$\begin{align} A&:=\int_0^2 \frac{\ln x}{2+x}\,dx\\ &=\frac{1}{2}\int_0^2 \frac{\ln x}{1+\frac{x}{2}}\,dx\\ \end{align}$

Perform the change of variable $\displaystyle y=\frac{x}{2}$,

$\begin{align} A&=\int_0^1 \frac{\ln(2y)}{1+y}\,dy\\ &=\int_0^1 \frac{\ln 2}{1+y}\,dy+\int_0^1 \frac{\ln y}{1+y}\,dy\\ &=\ln 2\Big[\ln(1+y)\Big]_0^1-\frac{\pi^2}{12}\\ &=\boxed{\ln^2 2-\frac{\pi^2}{12}} \end{align}$

It is well-known that,

$\begin{align}\int_0^1 \frac{\ln y}{1+y}\,dy=-\frac{1}{2}\zeta(2)\end{align}$

and,

$\displaystyle \zeta(2)=\frac{\pi^2}{6}$

$\begin{align}B&:=\int_0^2 \frac{\ln x}{x^2-2x+2}\,dx\\ &=\int_0^2\frac{\ln x}{(x-1)^2+1}\,dx \end{align}$

Perform the change of variable $y=x-1$,

$\begin{align}B&=\int_{-1}^1\frac{\ln(1+y)}{y^2+1}\,dy\\ &=\int_{-1}^0\frac{\ln(1+y)}{y^2+1}\,dy+\int_{0}^1\frac{\ln(1+y)}{y^2+1}\,dy \end{align}$

Perform the change of variable $x=-y$ in the first integral,

$\begin{align}B&=\int_{0}^1\frac{\ln(1-y)}{y^2+1}\,dy+\int_{0}^1\frac{\ln(1+y)}{y^2+1}\,dy\\ \end{align}$

In the first integral perform the change of variable $\displaystyle x=\frac{1-y}{1+y}$

$\begin{align}B&=\left(\int_0^1\frac{\ln 2}{1+x^2}\,dx+\int_0^1\frac{\ln x}{1+x^2}\,dx-\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx\right)+\int_{0}^1\frac{\ln(1+y)}{1+y^2}\,dy\\ &=\boxed{\frac{1}{4}\pi\ln 2-\text{G}} \end{align}$

It is well-known that,

$\displaystyle \int_0^1 \frac{\ln x}{1+x^2}\,dx=-\text{G}$

$\begin{align}C&:=\int_0^2 \frac{x\ln x}{x^2-2x+2}\,dx\\ &=\int_0^2\frac{x\ln x}{(x-1)^2+1}\,dx \end{align}$

Perform the change of variable $y=x-1$,

$\begin{align}C&=\int_{-1}^1\frac{(1+y)\ln(1+y)}{y^2+1}\,dy\\ &=B+\int_{-1}^1\frac{y\ln(1+y)}{y^2+1}\,dy\\ &=\frac{1}{4}\pi\ln 2-\text{G}+\int_{-1}^0\frac{y\ln(1+y)}{y^2+1}\,dy+\int_{0}^1\frac{y\ln(1+y)}{y^2+1}\,dy \end{align}$

In the first integral perform the change of variable $x=-y$,

$\begin{align}C&=\frac{1}{4}\pi\ln 2-\text{G}-\int_{0}^1\frac{x\ln(1-x)}{x^2+1}\,dx+\int_{0}^1\frac{x\ln(1+x)}{x^2+1}\,dx\\ &=\frac{1}{4}\pi\ln 2-\text{G}+\int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{1+x^2}\,dx \end{align}$

Perform the change of variable $\displaystyle y=\frac{1-x}{1+x}$,

$\begin{align}C&=\frac{1}{4}\pi\ln 2-\text{G}+\int_0^1 \frac{(x-1)\ln x}{x^3+x^2+x+1}\,dx\\ &=\frac{1}{4}\pi\ln 2-\text{G}+\int_0^1\frac{x\ln x}{1+x^2}\,dx-\int_0^1\frac{\ln x}{1+x}\,dx\\ &=\frac{1}{4}\pi\ln 2-\text{G}+\frac{1}{4}\int_0^1\frac{2x\ln(x^2)}{1+x^2}\,dx-\int_0^1\frac{\ln x}{1+x}\,dx\\ \end{align}$

In the first integral perform the change of variable $y=x^2$,

$\begin{align}C&=\frac{1}{4}\pi\ln 2-\text{G}+\frac{1}{4}\int_0^1\frac{\ln x}{1+x}\,dx-\int_0^1\frac{\ln x}{1+x}\,dx\\ &=\frac{1}{4}\pi\ln 2-\text{G}-\frac{3}{4}\int_0^1\frac{\ln x}{1+x}\,dx\\ &=\boxed{\frac{1}{4}\pi\ln 2-\text{G}+\frac{1}{16}\pi^2} \end{align}$

Therefore,

$\boxed{\displaystyle I=\frac{1}{10}\ln^2 2+\frac{3}{40}\pi\ln 2-\frac{3}{10}\text{G}-\frac{7}{480}\pi^2}$

Answer 2

Just for your curiosity.

If you enjoy special functions of complex arguments, the antiderivative can be computed.

$$\frac 1 {x^3-2x+4}=\frac{\frac1{10}}{ x+2}-\frac{\frac{1}{20}-\frac{3 i}{20}}{x-(1-i)}-\frac{\frac{1}{20}+\frac{3 i}{20}}{x-(1+i)}$$ which makes that we are left with integrals $$I_a=\int \frac{\log(x)}{x-a}\,dx=\text{Li}_2\left(\frac{x}{a}\right)+\log (x) \log \left(1-\frac{x}{a}\right)$$ where appears the polylogarithm function.

This would lead to the result Thomas Andrews gave in a comment.

Answer 3

To address your question of how to handle loops when integrating by parts, let $I=\int e^x\sin x \ dx$. Both functions are transcendental. We'll try using $u_1=e^x$ and $dv_1=\sin x \ dx$. These give $du_1=e^x \ dx$ and $v_1=-\cos x$. Thus, $$I=-e^x\cos x+\int e^x\cos x \ dx.$$ Now we have another integral. We'll try by parts again, and it's important we keep the same arrangement as last time (i.e. we need $u_2=du_1$ and $dv_2=v_1$; in this case we used exponential as $u$ and trigonometric as $v$, though we could have dove it the other way as long as we were consistent) otherwise we will just undo our last step. So, $u_2=e^x$ and $dv_2=\cos x \ dx$. These give $du_2=e^x \ dx$ and $v_2=\sin x$. Thus, $$ \begin{align} I&=-e^x\cos x+e^x\sin x-\int e^x\sin x \ dx \\ &=e^x(\sin x-\cos x)-I \\ 2I&=e^x(\sin x-\cos x) \\ I&=\frac{1}{2}e^x(\sin x-\cos x). \end{align} $$ The important step is realising that when you get back to where you started, you can perform algebra to solve for your result.

That said, I do not guarantee this is what's needed here, just thought it might be worth a try and then you asked. So here you go.