If for non-zero x, af(x) + bf(1/x) = 1/x - 25 where a != b, then int_1^2 f(x) dx is

A. (1/(a^2 - b^2)) * [a(2ln2 - 25) ...

Question

If for non-zero x, af(x) + bf(1/x) = 1/x - 25 where a != b, then int_1^2 f(x) dx is

A. (1/(a^2 - b^2)) * [a(2ln2 - 25) + b/2]
B. (1/(a^2 - b^2)) * [a(2ln2 - 25) - b/2]
C. (1/(a^2 - b^2)) * [a(2ln2 + 25) + b/2]
D. (1/(a^2 - b^2)) * [a(2ln2 - 25) + b]

Accepted Answer

(1/(a^2 - b^2)) * [a(2ln2 - 25) + b/2] We have two equations: af(x) + bf(1/x) = 1/x - 25. Replacing x by 1/x gives af(1/x) + bf(x) = x - 25. This is a 2x2 linear system in f(x) and f(1/x). Solving by elimination gives f(x) explicitly, then we integrate from 1 to 2.

If for non-zero x, af(x) + bf(1/x) = 1/x - 25 where a != b, then int_1^2 f(x) dx is A. (1/(a^2 - b^2)) * [a(2ln2 - 25) + b/2] B. (1/(a^2 - b^2)) * [a(2ln2 - 25) - b/2] C. (1/(a^2 - b^2)) * [a(2ln2 + 25) + b/2] D. (1/(a^2 - b^2)) * [a(2ln2 - 25) + b]