Let x neq frac{-3}{5}, frac{2}{5}. If fleft(f | vedclass

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(C) Given $f\left(\frac{2x+1}{5x+3}\right) = x+2$. Let $t = \frac{2x+1}{5x+3}$. Then $t(5x+3) = 2x+1 \implies 5xt + 3t = 2x+1 \implies x(5t-2) = 1-3t

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$\frac{7}{5}x - \frac{1}{25} \log |5x-2| + c$

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(C) Given $f\left(\frac{2x+1}{5x+3}\right) = x+2$. Let $t = \frac{2x+1}{5x+3}$. Then $t(5x+3) = 2x+1 \implies 5xt + 3t = 2x+1 \implies x(5t-2) = 1-3t \implies x = \frac{1-3t}{5t-2} = \frac{3t-1}{2-5t}$. Substituting $x$ into the function: $f(t) = \frac{3t-1}{2-5t} + 2 = \frac{3t-1 + 4-10t}{2-5t} = \frac{3-7t}{2-5t} = \frac{7t-3}{5t-2}$. Now, $\int f(x) dx = \int \frac{7x-3}{5x-2} dx$. Using division: $\frac{7x-3}{5x-2} = \frac{7}{5} \left(\frac{5x-2}{5x-2}\right) + \frac{\frac{14}{5}-3}{5x-2} = \frac{7}{5} - \frac{1}{5(5x-2)}$. Thus, $\int f(x) dx = \int \left(\frac{7}{5} - \frac{1}{5(5x-2)}\right) dx = \frac{7}{5}x - \frac{1}{25} \ln |5x-2| + c$.

Let $x \neq \frac{-3}{5}, \frac{2}{5}$. If $f\left(\frac{2x+1}{5x+3}\right) = x+2$, then $\int f(x) dx =$

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