$$\int \frac{\ln{(x^4 + x^2)}}{x^2} \mathrm{d}x$$
Can't solve this integral. I have been sitting over it for already an hour and still can't find an obvious solution.
Please help.
2026-04-24 03:50:31.1777002631
How to solve $\int \frac{\ln{(x^4 + x^2)}}{x^2} \mathrm{d}x$?
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$$\int\frac{\ln(x^2+x^4)}{x^2}dx=2\int\frac{\ln{x}}{x^2}dx+\int\frac{\ln(1+x^2)}{x^2}dx=2I_1+I_2$$ $I_1=\int\ln{x}(-x^{-1})'dx=-\frac{\ln{x}}{x}+\int\frac{1}{x^2}dx=-\frac{\ln{x}}{x}-\frac{1}{x}+C$.
$I_2=\int\ln(1+x^2)(-x^{-1})'dx=-\frac{\ln(1+x^2)}{x}+2\int\frac{1}{1+x^2}dx=2\arctan(x)-\frac{\ln(1+x^2)}{x}+C$