If fleft(frac{t+1}{2 t+1}right)=t+1, then int | vedclass

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(D) We have, $f\left(\frac{t+1}{2 t+1}\right)=t+1$. Let $\frac{t+1}{2 t+1}=x$. Then $t+1=x(2t+1) \Rightarrow t+1=2tx+x$. $t(1-2x)=x-1 \Rightarrow t=\f

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$\frac{x}{2}+\frac{1}{4} \log (2 x-1)+c$

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(D) We have, $f\left(\frac{t+1}{2 t+1}\right)=t+1$. Let $\frac{t+1}{2 t+1}=x$. Then $t+1=x(2t+1) \Rightarrow t+1=2tx+x$. $t(1-2x)=x-1 \Rightarrow t=\frac{x-1}{1-2x}=\frac{1-x}{2x-1}$. Substituting $t$ into the expression for $f(x)$: $f(x) = \frac{1-x}{2x-1} + 1 = \frac{1-x+2x-1}{2x-1} = \frac{x}{2x-1}$. Now, we calculate the integral: $\int f(x) dx = \int \frac{x}{2x-1} dx = \frac{1}{2} \int \frac{2x}{2x-1} dx$. $= \frac{1}{2} \int \frac{2x-1+1}{2x-1} dx = \frac{1}{2} \int \left(1 + \frac{1}{2x-1}\right) dx$. $= \frac{1}{2} \left[ x + \frac{1}{2} \log |2x-1| \right] + c$. $= \frac{x}{2} + \frac{1}{4} \log |2x-1| + c$.

If $f\left(\frac{t+1}{2 t+1}\right)=t+1$, then $\int f(x) d x=$

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