Let 5 f(x)+4 fleft(frac{1}{x}right)=frac{1}{x | vedclass

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(A) Given the equation: $5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3$  $(1)$Replace $x$ with $\frac{1}{x}$ in equation $(1)$:$5 f\left(\frac{1}{x

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$10 \log _e 2-6$

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(A) Given the equation: $5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3$  $(1)$Replace $x$ with $\frac{1}{x}$ in equation $(1)$:$5 f\left(\frac{1}{x}\right)+4 f(x)=x+3$  $(2)$To eliminate $f\left(\frac{1}{x}\right)$,multiply $(1)$ by $5$ and $(2)$ by $4$:$25 f(x)+20 f\left(\frac{1}{x}\right)=\frac{5}{x}+15$$16 f(x)+20 f\left(\frac{1}{x}\right)=4x+12$Subtract the second equation from the first:$9 f(x) = \frac{5}{x} - 4x + 3$$f(x) = \frac{1}{9} \left( \frac{5}{x} - 4x + 3 \right)$Now,calculate the integral $I = 18 \int \limits_1^2 f(x) \, dx$:$I = 18 \int \limits_1^2 \frac{1}{9} \left( \frac{5}{x} - 4x + 3 \right) \, dx$$I = 2 \int \limits_1^2 \left( \frac{5}{x} - 4x + 3 \right) \, dx$$I = 2 \left[ 5 \ln|x| - 2x^2 + 3x \right]_1^2$$I = 2 \left[ (5 \ln 2 - 2(4) + 3(2)) - (5 \ln 1 - 2(1) + 3(1)) \right]$$I = 2 \left[ (5 \ln 2 - 8 + 6) - (0 - 2 + 3) \right]$$I = 2 \left[ 5 \ln 2 - 2 - 1 \right]$$I = 2 \left[ 5 \ln 2 - 3 \right] = 10 \ln 2 - 6$

Let $5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3$,where $x > 0$. Then $18 \int \limits_1^2 f(x) \, dx$ is equal to:

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