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One of an exam's task was to calculate the following integral: $$I=\int_{0}^{\infty}\frac{x^3\ln x}{(x^4+1)^3} \,dx$$ I tried integration by parts: $$I=\frac{1}{4}\int_{0}^{\infty}\ln x \cdot (x^4+1)^{-3} \,d(x^4+1)$$ but then things got messy and I ultimately concluded that the improper integral diverges, which is wrong since the answer is: $$I=-\frac{1}{32}$$ What should I do?

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  • $\begingroup$ How did you "prove" it? :) $\endgroup$ – Vincenzo Oliva Sep 5 '15 at 8:31
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    $\begingroup$ Try the substitution $u=x^4+1$ $\endgroup$ – uniquesolution Sep 5 '15 at 8:31
  • $\begingroup$ How will it help for the expression inside $\ln$? @uniquesolution $\endgroup$ – Aditya Agarwal Sep 5 '15 at 8:55
  • $\begingroup$ @Aditya Agarwal it will be a first step to a possible solution. Not everything is instantly visible at all times. $\endgroup$ – uniquesolution Sep 5 '15 at 8:59
  • $\begingroup$ It won't, it would only complicate the things more. $\endgroup$ – Aditya Agarwal Sep 5 '15 at 8:59
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Integrating by parts, $$ \int\frac{x^3\log x}{(1+x^4)^3}\,dx=-\frac{\log x}{8(1+x^4)^2}+\int \frac{1}{8(1+x^4)^2x}\,dx $$ Doing partial fraction decomposition, $$ \frac{1}{8(1+x^4)^2x}=\frac{1}{8x}-\frac{x^3}{8(1+x^4)}-\frac{x^3}{8(1+x^4)^2}. $$

Can you take it from here?

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  • $\begingroup$ That seems the right way to do it which is the same with my post, but I didn't proceed further with this because this type of decomposition didn't come to me at the time. $\endgroup$ – user209217 Sep 5 '15 at 10:30
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$$I = \int\frac{x^3\ln(x)}{(x^4+1)^3}dx=\frac{\ln x}4\int \frac{4x^3}{(x^4+1)^3}dx-\int\left(\frac1{4x}\int \frac{4x^3}{(x^4+1)^3}dx\right)dx $$ $$=-\frac{\ln x}{8(x^4+1)^2}+\frac 1{32}\int \frac{4x^3}{x^4(x^4+1)^2}dx$$ $$=-\frac{\ln x}{8(x^4+1)^2}+\frac1{32}\ln\left(\frac{x^4}{1+x^4}\right)+\frac 1{32(1+x^4)}$$

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Given $$\displaystyle I = \int_{0}^{\infty}\frac{x^3\ln(x)}{(x^4+1)^3}dx\;,$$ Now Let $\displaystyle x^2 =\tan \phi\;,$ Then $2x dx = \sec^2 \phi d\phi$ and changing limit

We get $$\displaystyle I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}}\frac{\tan \phi \cdot \sec^2 \phi }{\sec^6 \phi}\cdot \ln (\tan \phi)d\phi = \frac{1}{4}\int_{0}^{\frac{\pi}{2}}\cos^3 \phi \cdot \sin \phi \cdot \ln(\tan \phi)d\phi$$

Now Using $$\displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$$

So we get $$\displaystyle I = -\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\cos \phi \cdot \sin^3 \phi\cdot \ln(\tan )\phi d\phi$$

Now $$\displaystyle 2I= \frac{1}{8}\int_{0}^{\frac{\pi}{2}}\sin 2\phi \cdot \cos 2\phi \ln (\tan \phi)d\phi = \frac{1}{16}\int_{0}^{\frac{\pi}{2}}\sin (4\phi)\cdot \ln(\tan \phi )d\phi$$

So we got $$\displaystyle I = \frac{1}{32}\int_{0}^{\frac{\pi}{2}}\sin (4\phi)\cdot \ln(\tan \phi) d\phi = \frac{2}{32}\int_{0}^{\frac{\pi}{4}}\sin (4\phi)\cdot \ln(\tan \phi) d\phi$$

Now Using Integration by parts,

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