Evaluating $\\int_0^{\\infty}\\frac{\\ln(x^2+1)}{x^2+1}dx$ How would I go about evaluating this integral? $$\int_0^ {\infty}\frac {\ln (x^2+1)} {x^2+1}dx $$ What I've tried so far: I tried a semicircular integral in the positive imaginary part of the complex p
algebra precalculus - Evaluating $\frac {1} {a^ {2025}}+\frac {1} {b . . . When I tried to solve this problem, I found a solution (official) video on YouTube That is a = −b, c = 2024 a = b, c = 2024 and the correct answer is 1 20242025 1 2024 2025 Is there an alternative solution but not using (a + b)(a + c)(b + c) + abc = (a + b + c)(ab + ac + bc) (a + b) (a + c) (b + c) + a b c = (a + b + c) (a b + a c + b c) ?