જો gleft(frac{t+1}{2 t+1}right)=t+1 હોય,તો in | vedclass

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(D) આપેલ છે કે $g\left(\frac{t+1}{2 t+1}\right)=t+1$. ધારો કે $x = \frac{t+1}{2 t+1}$. તેથી $x(2t+1) = t+1 \Rightarrow 2tx + x = t+1 \Rightarrow t(2x-

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$\frac{x}{2}+\frac{1}{4} \log _e|2 x-1|+c$

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(D) આપેલ છે કે $g\left(\frac{t+1}{2 t+1}\right)=t+1$. ધારો કે $x = \frac{t+1}{2 t+1}$. તેથી $x(2t+1) = t+1 \Rightarrow 2tx + x = t+1 \Rightarrow t(2x-1) = 1-x$. આમ,$t = \frac{1-x}{2x-1}$. $g(x) = t+1$ માં $t$ ની કિંમત મૂકતા: $g(x) = \frac{1-x}{2x-1} + 1 = \frac{1-x+2x-1}{2x-1} = \frac{x}{2x-1}$. હવે,$\int g(x) dx = \int \frac{x}{2x-1} dx$. $= \frac{1}{2} \int \frac{2x}{2x-1} dx = \frac{1}{2} \int \frac{2x-1+1}{2x-1} dx$. $= \frac{1}{2} \int \left(1 + \frac{1}{2x-1}\right) dx$. $= \frac{1}{2} \left(x + \frac{1}{2} \log _e|2x-1|\right) + C$. $= \frac{x}{2} + \frac{1}{4} \log _e|2x-1| + C$.

જો $g\left(\frac{t+1}{2 t+1}\right)=t+1$ હોય,તો $\int g(x) d x=$

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