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) ?
Evaluating $\\lim\\limits_{R\\to +∞}\\iint_{x^2+y^2\\leq R^2}\\left . . . I wonder whether you would agree that the second line above is easier to read than the first Note (1) the use of \left and \right, which makes the parentheses assume appropriate sizes, (2) the use of \limits, which affects the position of the bounds of integration, and (3) small spaces separating dx d x and dy d y from what precedes and follows them