Evaluate $\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx$

Question

$$\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx $$ Help me please. I don't know any ways of solution. Thank you.

Answer 1

Integrating by Parts,

$$I=\int \ln x\cdot\frac x{(1+x^2)^2}dx$$

$$=\ln x\int\frac x{(1+x^2)^2}dx-\int\left(\frac{d(\ln x)}{dx}\cdot \frac x{(1+x^2)^2}dx\right)dx $$

For $\displaystyle J=\int\frac x{(1+x^2)^2}dx$ set $1+x^2=u\implies2x\ dx=du$

$\displaystyle J=\int\frac{du}{2u^2}=-\frac1{2u}=-\frac1{2(1+x^2)}$

$\displaystyle\implies I=\ln x\left(-\frac1{2(1+x^2)}\right)+\int\frac1{2x(1+x^2)}\ dx$

Setting $x^2=v,$ $\displaystyle K=\int\frac1{2x(1+x^2)}\ dx=\int\frac{2x}{4x^2(1+x^2)}\ dx$

$\displaystyle4K=\int\frac{dv}{v(1+v)}=\int\frac{v+1-v}{v(1+v)}dv=\int\frac{dv}v -\int\frac{dv}{v+1}=\cdots$

Answer 2

Another approach:

expand $\frac{1}{(1+x^2)^2} = \sum_{k=0}^{\infty}(-1)^kkx^{2k}$, then each integral becomes of the form $$ I_k = (-1)^kk\int_{0}^{1}x^{2k+1} \log x dx $$ and integrate by parts $du = x^{2 k +1}, v = \log x$. Lots of things will cancel out because of upper and lower bounds on the integral.

Answer 3

Rewrite: $$ \int\frac{x\ln x}{(1+x^2)^2}\ dx=\frac14\int\frac{2x}{1+x^2}\cdot\ln x\cdot\frac{2\ dx}{1+x^2}. $$ Now, consider Weierstrass substitution: $x=\tan\dfrac t2$, $\sin t=\dfrac{2x}{1+x^2}$, $\cos t=\dfrac{1-x^2}{1+x^2}$, and $dt=\dfrac{2\ dx}{1+x^2}$. The integral turns out to be $$ \frac14\int\sin t\cdot \ln \tan\frac t2\ dt. $$ Use IBP by taking $u=\ln \tan\dfrac t2\;\Rightarrow\;du=\dfrac12\dfrac{\sec^2\frac t2}{\tan\frac t2}\ dt=\dfrac{dt}{\sin t}$ and $dv=\sin t\ dt\;\Rightarrow\;v=-\cos t$. We obtain \begin{align} \frac14\int\sin t\cdot \ln \tan\frac t2\ dt&=\frac14\left[-\cos t\cdot\ln \tan\dfrac t2+\int\dfrac{\cos t}{\sin t}\ dt\right]\\ &=\frac14\left[-\cos t\cdot\ln \tan\dfrac t2+\int\dfrac{1}{\sin t}\ d(\sin t)\right]\\ &=\frac14\left[-\cos t\cdot\ln \tan\dfrac t2+\ln\sin t\right]+C\\ &=\frac14\left[-\dfrac{1-x^2}{1+x^2}\cdot\ln x+\ln\dfrac{2x}{1+x^2}\right]+C\\ &=\frac14\left[\dfrac{x^2\ln x-\ln x}{1+x^2}+\ln2+\ln x-\ln\left(1+x^2\right)\right]+C\\ &=\frac14\left[\dfrac{x^2\ln x-\ln x+\left(1+x^2\right)\ln x}{1+x^2}+\ln2-\ln\left(1+x^2\right)\right]+C\\ &=\frac14\left[\dfrac{2x^2\ln x}{1+x^2}-\ln\left(1+x^2\right)\right]+\frac{\ln2}{4}+C\\ \int\frac{x\ln x}{(1+x^2)^2}\ dx&=\frac14\left[\dfrac{2x^2\ln x}{1+x^2}-\ln\left(1+x^2\right)\right]+K. \end{align} Thus $$ \int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx=\frac14\left[\dfrac{2x^2\ln x}{1+x^2}-\ln\left(1+x^2\right)\right]_{x=0}^1=\large\color{blue}{-\frac{\ln2}{4}}. $$

Answer 4

First consider $$ I(s) = \int_0^1\frac{x^{s}}{(1+ x^2)^2} \ \mathrm{d}x, $$ Then \begin{align} I(s) & = \int_0^1 x^s \displaystyle \sum_{n=1}^{\infty} n(-1)^{n+1}x^{2n-2} \ \mathrm{d}x, \\ & = \displaystyle \sum_{n=1}^{\infty} \frac{n(-1)^{n+1}}{2n+s-1}, \end{align} And our original integral is $$ \frac{\partial I }{\partial s}\bigg|_{s=1} = -\displaystyle \sum_{n=1}^{\infty} \frac{n(-1)^{n+1}}{(2n)^2} = -\frac{1}{4}\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = -\frac{\ln 2}{4}. $$ Edit - This is simply Alex's approach. I took an unnecessary detour.

Answer 5

For $x\ge0$

$$I=\int\frac{x\ln x}{(1+x^2)^2}\ dx=\frac12\int\frac{x\ln(x^2)}{(1+x^2)^2}\ dx$$

Setting $\displaystyle x^2=u,4I=\int\frac{\ln y}{(1+y)^2}dy$

Integrating by Parts, $\displaystyle4I=\ln y\int\frac{dy}{(1+y)^2}-\int\left(\frac{d(\ln y)}{dy}\cdot\int\frac{dy}{(1+y)^2}\right)dy$

$\displaystyle4I=-\frac{\ln y}{1+y}+\int\frac{dy}{y(1+y)}$ which can be solved easily by using my other answer

Answer 6

If I may pose yet another method to go about this one.

Make the sub $x=e^{-t}$

$$-\int_{0}^{\infty}\frac{te^{-2t}}{(e^{-2t}+1)^{2}}dt=-1/4\int_{0}^{\infty}\frac{t}{\cosh^{2}(t)}dt$$

Integrate by parts: $\displaystyle u=t, \;\ dv=\frac{1}{\cosh^{2}(t)}dt$

resulting in:

$$-1/4\lim_{t\to \infty}\left[t\cdot \tanh(t)-\log(\cosh(t))\right]$$

Rewrite as $$\lim_{t\to \infty}\frac{t\cdot \sinh(t)+\log(sech(t))\cosh(t)}{\cosh(t)}$$

and apply L'Hopitals rule:

This gives:

$$\lim_{t\to \infty}-log(1+e^{2t})-\frac{2t}{e^{2t}+1}+\frac{4}{e^{2t}+1}-\frac{4}{(e^{2t}+1)^{2}}+2t+\log(2)$$

Note that as $t\to \infty$, the $2t$ and the $-log(1+e^{2t})$ cancel one another. All other terms go to 0 except for $\log(2)$

This results in our final answer of $$-1/4 \log(2)$$

This may be a fun one to try using contours.

Answer 7

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{x\ln\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x:\ {\large ?}}$

With $\ds{\large 0 < \epsilon < 1}$: \begin{align} &\color{#c00000}{\int_{\epsilon}^{1}{x\ln\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x} =-\,\half\int_{x = \epsilon}^{x = 1}\ln\pars{x}\dd\pars{1 \over 1 + x^{2}} =\half\,{\ln\pars{\epsilon} \over 1 + \epsilon^{2}} +\half\int_{\epsilon}^{1}{1 \over x^{2} + 1}\,{1 \over x}\,\dd x \\[3mm]&=\half\,{\ln\pars{\epsilon} \over 1 + \epsilon^{2}} +\half\Im\int_{\epsilon}^{1}{1 \over x - \ic}\,{1 \over x}\,\dd x =\half\,{\ln\pars{\epsilon} \over 1 + \epsilon^{2}} -\half\Re\int_{\epsilon}^{1}\pars{{1 \over x - \ic} - {1 \over x}}\,\dd x \\[3mm]&=\half\,{\ln\pars{\epsilon} \over 1 + \epsilon^{2}} -\half\Re\ln\pars{\bracks{1 - \ic}/1 \over \bracks{\epsilon - \ic}/\epsilon} =\half\,{\ln\pars{\epsilon} \over 1 + \epsilon^{2}} -\half\Re \ln\pars{\epsilon\,{1 + \epsilon + \bracks{1 - \epsilon}\ic \over \epsilon^{2} + 1}} \\[3mm]&=\half\,{\ln\pars{\epsilon} \over 1 + \epsilon^{2}}-\half \ln\pars{\epsilon\,{\root{2 + 2\epsilon^{2}} \over \epsilon^{2} + 1}} \\[3mm]&=\underbrace{\half\,\ln\pars{\epsilon}\pars{{1 \over 1 + \epsilon^{2}} - 1}} _{\to\ 0\quad\mbox{when}\quad \epsilon \to 0^{+}}\ -\ {1 \over 4}\,\ln\pars{2}\ +\ \underbrace{{1 \over 4}\ln\pars{1 + \epsilon^{2}}}_{\to\ 0\quad\mbox{when}\quad \epsilon \to 0^{+}} \end{align}

$$\color{#00f}{\large% \int_{0}^{1}{x\ln\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x =-\,{1 \over 4}\,\ln\pars{2}} $$

Answer 8

Integration by parts with $u=\ln x$, $dv=xdx/(1+x^2)^2$, followed by partial fractions gives the indefinite integral

$$\begin{align} \int{x\ln x\over(1+x^2)^2}dx&={-\ln x\over2(1+x^2)}+{1\over2}\int{1\over x(1+x^2)}dx\\ &={-\ln x\over2(1+x^2)}+{1\over2}\int\left({1\over x}-{x\over(1+x^2)}\right)dx\\ &={-\ln x\over2(1+x^2)}+{1\over2}\left(\ln x-{1\over2}\ln(1+x^2) \right)\\ &={x^2\ln x\over2(1+x^2)}-{1\over4}\ln(1+x^2) \end{align}$$

When you turn this into a definite integral from $0$ to $1$, the term with $x^2\ln x$ in the numerator is $0$ at both limits, so you're left with

$$\int_0^1{x\ln x\over(1+x^2)^2}dx=-{1\over4}\ln2$$

Note, you can't just write

$$\int_0^1{x\ln x\over(1+x^2)^2}dx={-\ln x\over2(1+x^2)}\Big|_0^1+{1\over2}\int_0^1{1\over x(1+x^2)}dx$$

because the two terms on the right hand side there diverge at $x=0$. For this problem, you really have to find the complete antiderivative before you plug in limits.