<s>[INST] Calculation of improper integral: $\int_{0}^{\infty}\frac{x^3\ln x}{(x^4+1)^3} \,dx$ 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?
 [/INST] $$I = \int\frac{x^3\ln(x)}{(x^4+1)^3}dx=\frac{\ln x}4\int \frac{4x^3}{(x^4+1)^3}dx-\int\left(\frac1{4x}\int \frac{4x^3}{(x^4+1)^3}dx\right)dx $$
$$=-\frac{\ln x}{8(x^4+1)^2}+\frac 1{32}\int \frac{4x^3}{x^4(x^4+1)^2}dx$$
$$=-\frac{\ln x}{8(x^4+1)^2}+\frac1{32}\ln\left(\frac{x^4}{1+x^4}\right)+\frac 1{32(1+x^4)}$$
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