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Question Number 152350 by mnjuly1970 last updated on 27/Aug/21

$$$$$$\mathrm{prove}\mathrm{that}....$$$$$$$$\mathit{\varphi }:={\int }_0^1\frac{x.{ln}^2(1+x)}{1+x^2}dx=\frac{1}{96}ln\left(2\right)({\pi }^2+4{ln}^2\left(2\right))....◼$$$$\mathrm{M}.\mathrm{N}..$$$$$$$$$$

Answered by mindispower last updated on 27/Aug/21

$$x\to \frac{1-t}{1+t}\Rightarrow dx=-\frac{2dt}{{(1+t)}^2}$$$$$$$$\iff \varphi ={\int }_0^1\frac{\frac{1-t}{1+t}{ln}^2\left(\frac{2}{1+t}\right)}{1+t^2}.dt$$$${\int }_0^1\frac{1-t}{(1+t)(1+t^2)}({ln}^2\left(2\right)+{ln}^2(1+t)-2ln\left(2\right)ln(1+t))dt$$$$\frac{1-t}{(1+t)(1+t^2)}=\frac{1}{1+t}-\frac{t}{1+t^2}$$$$={\int }_0^1(\frac{1}{1+t}-\frac{t}{1+t^2})({ln}^2\left(2\right)+{ln}^2(1+t)-2ln\left(2\right)ln(1+t))dt$$$$={ln}^2\left(2\right){\int }_0^1(\frac{1}{1+t}-\frac{t}{1+t^2})dt+{\int }_0^1\frac{{ln}^2(1+t)}{1+t}-\varphi$$$$-2ln\left(2\right){\int }_0^1\frac{ln(1+t)}{1+t}dt+2ln\left(2\right){\int }_0^1\frac{tln(1+t)}{1+t^2}dt$$$$2\varphi ={ln}^2\left(2\right)\left[\frac{ln\left(2\right)}{2}\right]+\frac{1}{3}{ln}^3\left(2\right)-2ln\left(2\right).\frac{{ln}^2\left(2\right)}{2}$$$$+2ln\left(2\right){\int }_0^1\frac{tln(1+t)}{1+t^2}dt$$$${\int }_0^1\frac{tln(1+t)}{1+t^2}=S$$$$f\left(a\right)={\int }_0^1\frac{tln(1+at)}{1+t^2}dt$$$$S={\int }_0^1{\int }_0^1\frac{t^{2}dt}{(1+t^2)(1+at)}da$$$$={\int }_0^1\frac{a(aln\left(4\right)-\pi )+4ln(a+1)}{4(a^3+a)}$$$$={\int }_0^1\frac{aln\left(4\right)-\pi }{4(1+a^2)}da+{\int }_0^1\frac{ln(1+a)}{a+a^3}da$$$$=\frac{ln\left(4\right)}{8}ln\left(2\right)-\frac{{\pi }^2}{16}+{\int }_0^1\frac{ln(1+a)}{a}-\frac{aln(1+a)}{1+a^2}da$$$$=\frac{{ln}^2\left(2\right)}{4}-\frac{{\pi }^2}{16}+{\int }_0^1\frac{ln(1-(-a))}{-a}d(-a)-S$$$$S=\frac{{ln}^2\left(2\right)}{8}-\frac{{\pi }^2}{32}-\frac{1}{2}{li}_2(-1)$$$$=\frac{{ln}^2\left(2\right)}{8}+\frac{{\pi }^2}{96}$$$$2\varphi =-\frac{{ln}^3(2}{6}+2ln\left(2\right)(\frac{{ln}^2\left(2\right)}{8}+\frac{{\pi }^2}{96})$$$$=-\frac{{ln}^3\left(2\right)}{6}+\frac{{ln}^3\left(2\right)}{4}+\frac{{\pi }^{2}ln\left(2\right)}{48}=\frac{{ln}^3\left(2\right)}{12}+\frac{ln\left(2\right)}{48}{\pi }^2$$$$=\frac{ln\left(2\right)}{48}({\pi }^2+4{ln}^2\left(2\right))$$$$\varphi =\frac{ln\left(2\right)}{96}({\pi }^2+4{ln}^2\left(2\right))$$

Commented by mindispower last updated on 29/Aug/21

$$thanxwithepleasur$$

Commented by mnjuly1970 last updated on 28/Aug/21

$$verynicesirpower...grateful$$

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