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How to solve this definite integral? It cannot be solved using simple integration by parts.

$$\int_0^1 \frac{x}{1+x^2} \ln{(x+1)}\ dx$$ I encountered this in Issac Physics, so it should be solved just using high school calculus knowledge.

Quanto
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Ryan Hu
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    Have you tried anything? Any thoughts, do share it here! – Soumil Gupta Dec 07 '21 at 13:00
  • Which $u,,v$ did you try in integration by parts? Show us what happened. – J.G. Dec 07 '21 at 13:12
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    You'll find another technique more useful. Evaluate$$\int_0^1\left[\frac{d}{dc}\left(\int_0^1\frac{x}{1+x^2}\ln(1+cx)dx\right)\right]dc.$$ – J.G. Dec 07 '21 at 13:24
  • Basically when I first saw it I tried integration by parts(that's what you usually do when there is $\ln{f(x)}$ right?). But if you do $u = \ln{(x+1)}$ and $v' = \frac{x}{1+x^2}$ then you would see that another $\ln{f(x)}$ comes out as $\int \frac{x}{1+x^2}\ dx$ evaluates to $\frac{1}{2} \ln{(1+x^2)}$ – Ryan Hu Dec 07 '21 at 13:55
  • @J.G. That seems confusing, differentiating a definite integral should give a result of $0$ isn't it? – Ryan Hu Dec 07 '21 at 14:00
  • There are various techniques applicable here: double integrals, series, differentiation under the integral sign, etc. But I don't think there is a "high school calculus" solution involving simple substitutions or integration by parts. This integral is more advanced. – bjorn93 Dec 07 '21 at 14:15
  • @RyanHu The inner integral over $x$ has a $c$-dependent integrand, so the integral's $c$ derivative is nonzero. In fact, that derivative is$$\int_0^1\frac{x^2dx}{(1+cx)(1+x^2)},$$which you can evaluate by partial fractions. The resulting function of $c$ needs to be integrated from $0$ to $1$. – J.G. Dec 07 '21 at 14:19
  • @bjorn93 Oh, that's unfortunate, thanks for telling me tho. – Ryan Hu Dec 07 '21 at 14:25
  • @J.G. Thank you! I'll try – Ryan Hu Dec 07 '21 at 14:37
  • A further hint at @bjorn93's suggested complexity: the answer is $\frac{\pi^2+12\ln^22}{96}$, so an integral representation of $\zeta(2)$ will come up. – J.G. Dec 07 '21 at 14:40
  • See https://math.stackexchange.com/questions/4030962/calculating-int-01-fracx-lnx1x21-dx-without-using-complex-numbers?noredirect=1 – FDP Sep 26 '24 at 11:25

4 Answers4

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Using integration by parts, we can get $$\begin{align} \int_0^1{\frac{x}{x^2+1}\ln \left( 1+x \right) \mathrm{d}x}=&\frac{1}{2}\int_0^1{\frac{1}{x^2+1}\ln \left( 1+x \right) \mathrm{d}\left( x^2+1 \right)} \\ =&\frac{1}{2}\int_0^1{\ln \left( 1+x \right) \mathrm{d}\ln \left( x^2+1 \right)} \\ =&\frac{1}{2}\left( \ln ^22-\int_0^1{\frac{\ln \left( 1+x^2 \right)}{1+x}\mathrm{d}x} \right) \end{align}$$ And the second integral $\int_0^1{\frac{\ln \left( 1+x^2 \right)}{1+x}\mathrm{d}x}$ can be found in this cite, for example, from here we can get that $$ \int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}\mathrm{d}x=\frac{3}{4}\ln^{2}(2)-\frac{{\pi}^{2}}{48} $$ So combining them, we can get $$\begin{align} \int_0^1{\frac{x}{x^2+1}\ln \left( 1+x \right) \mathrm{d}x}=&\frac{1}{2}\left( \ln ^22-\left( \frac{3}{4}\ln ^22-\frac{\pi ^2}{48} \right) \right) \\ =&\frac{\pi ^2}{96}+\frac{1}{8}\ln ^22 \end{align}$$

Soriak
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  • @RyanHu What's more, I am curious about the range of your so called "high school calculus knowledge", as we are in the different provinces(seen from your profile) for college entrance examination. Knowledge such as integration by parts and variable substitution are beyond what we had to learn in high school. – Soriak Dec 08 '21 at 07:25
  • The integration by parts here simply swaps one integral with another integral of roughly the same complexity. It doesn't simplify the problem. Notice that the technique used in solving the second integral can be applied directly to the given integral. – bjorn93 Dec 08 '21 at 15:50
  • @SHBooKP By high school calculus knowledge, I mean what you would expect from a AP calculus BC student or an a level further math student(Im the latter one). Cuz I'm in an international school. – Ryan Hu Dec 14 '21 at 12:37
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No double integral or derivating parametric integral here. \begin{align}J&=\int_0^1\frac{x\ln(1+x)}{1+x^2}dx\\ K&=\int_0^1\frac{x\ln(1-x)}{1+x^2}dx\\ J+K&=\int_0^1\frac{x\ln(1-x^2)}{1+x^2}dx\\ &\overset{u=x^2}=\frac{1}{2}\int_0^1\frac{\ln(1-u)}{1+u}du\\ &\overset{z=\frac{1-u}{1+u}}=\frac12\int_0^1\frac{\ln\left(\frac{2z}{1+z}\right)}{1+z}du\\ &=\frac12\int_0^1\frac{\ln z}{1+z}dz+\frac{\ln^22}{4}\\ K-J&=\int_0^1\frac{x\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx\\ &\overset{u=\frac{1-x}{1+x}}=\int_0^1\frac{\ln u}{1+u}du-\underbrace{\int_0^1\frac{u\ln u}{1+u^2}du}_{=z=u^2}\\ &=\frac{3}{4}\int_0^1\frac{\ln u}{1+u}du \end{align} Therefore, \begin{align}J&=\frac{1}{2}\Big(\left(J+K\right)-\left(K-J\right)\Big)=\frac{\ln^22}{8}-\frac{1}{8}\int_0^1\frac{\ln u}{1+u}du\\ &\boxed{J=\frac{\ln^22}{8}+\frac{\pi^2}{96}} \end{align} \begin{align}K&=\frac{1}{2}\Big(\left(J+K\right)+\left(K-J\right)\Big)=\frac{\ln^22}{8}+\frac{5}{8}\int_0^1\frac{\ln u}{1+u}du\\ &\boxed{K=\frac{\ln^22}{8}-\frac{5\pi^2}{96}} \end{align}

NB: I assume, \begin{align}\int_0^1\frac{\ln x}{1+x}dx=-\frac{\zeta(2)}{2}=-\frac{\pi^2}{12}\end{align}

FDP
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$$\begin{align*}I=&\int_0^1{\frac{x\ln \left( 1+x \right)}{1+x^2}\mathrm{d}x}\\=&\int_0^1{\int_0^1{\frac{x^2}{\left( 1+x^2 \right) \left( 1+xt \right)}}\mathrm{d}t\mathrm{d}x}\\=&\underset{I_1}{\underbrace{\int_0^1{\int_0^1{\frac{1}{1+xt}}\mathrm{d}t\mathrm{d}x}}}-\int_0^1{\int_0^1{\frac{1}{\left( 1+x^2 \right) \left( 1+xt \right)}}\mathrm{d}t\mathrm{d}x}\\=&I_1+\underset{I_2}{\underbrace{\int_0^1{\int_0^1{\frac{tx-1}{\left( 1+x^2 \right) \left( 1+t^2 \right)}}\mathrm{d}t\mathrm{d}x}}}-\underset{I}{\underbrace{\int_0^1{\int_0^1{\frac{t^2}{\left( 1+t^2 \right) \left( 1+tx \right)}}\mathrm{d}t\mathrm{d}x}}}\end{align*}$$

Therefore $I=\frac{I_1+I_2}{2}$, where

$$\begin{align*}I_1=\int_0^1{\int_0^1{\frac{1}{1+xt}}\mathrm{d}t\mathrm{d}x}=&\int_0^1{\frac{\ln \left( 1+x \right)}{x}\mathrm{d}x}\\=&\int_0^1{\frac{1}{x}\sum_{n=1}^{+\infty}{-\frac{\left( -x \right) ^n}{n}}}\mathrm{d}x\\=&\sum_{n=0}^{+\infty}{\frac{\left( -1 \right) ^n}{n+1}\int_0^1{x^n\mathrm{d}x}}\\=&\sum_{n=0}^{+\infty}{\frac{\left( -1 \right) ^n}{\left( n+1 \right) ^2}}=\frac{\pi ^2}{6}\left( 1-2\cdot \frac{1}{4} \right) \\=&\frac{\pi ^2}{12}\end{align*}$$

and

$$\begin{align*}I_2=&\int_0^1{\int_0^1{\frac{tx-1}{\left( 1+x^2 \right) \left( 1+t^2 \right)}}\mathrm{d}t\mathrm{d}x}\\=&\int_0^1{\int_0^1{\frac{tx}{\left( 1+x^2 \right) \left( 1+t^2 \right)}}\mathrm{d}t\mathrm{d}x}-\int_0^1{\int_0^1{\frac{1}{\left( 1+x^2 \right) \left( 1+t^2 \right)}}\mathrm{d}t\mathrm{d}x}\\=&\left( \int_0^1{\frac{t}{1+t^2}}\mathrm{d}t \right) ^2-\left( \int_0^1{\frac{1}{1+t^2}}\mathrm{d}t \right) ^2\\=&\left( \frac{\ln 2}{2} \right) ^2-\left( \frac{\pi}{4} \right) ^2\end{align*}$$ Finally $$\begin{align*}\therefore I=&\frac{1}{2}\left( \frac{\pi ^2}{12}+\left( \frac{\ln 2}{2} \right) ^2-\left( \frac{\pi}{4} \right) ^2 \right) \\=&\frac{\pi ^2}{96}+\frac{\ln ^22}{8}\end{align*}$$

Soriak
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\begin{align} I=&\int_0^1 \frac{x \ln{(x+1)}}{1+x^2}\ dx =\int_0^1 \int_0^1 \frac{x^2}{(1+x^2)(1+x y)}dy\ dx\\ =& -I+\int_0^1 \frac{\ln(1+y)}y+\frac{2y\ln2-\pi }{4(1+y^2)}\ dy =\frac{1}{8}\ln ^22+ \frac{\pi ^2}{96} \end{align} where $\int_0^1 \frac{\ln(1+y)}ydy =\frac{\pi^2}{12}$.

Quanto
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