If $g\left(\frac{t+1}{2 t+1}\right)=t+1$, then $\int g(x) d x=$
If $g\left(\frac{t+1}{2 t+1}\right)=t+1$, then $\int g(x) d x=$
- $\frac{x^2}{2}+c$
- $\log _e(2 x-1)+\frac{1}{2} \log _e|(x+1)|+c$
- $\frac{1}{2} \log _e\left|\left(\frac{x+1}{2 x+1}\right)\right|+c$
- $\frac{x}{2}+\frac{1}{4} \log _e|2 x-1|+c$
Solution
$g\left(\frac{t+1}{2 t+1}\right)=t+1$
replacing $\frac{t+1}{2 t+1}$ by $x$, we get,
$
\begin{gathered}
\frac{t+1}{2 t+1}=x \Rightarrow t+1=2 t x+x \\
t(1-2 x)=x-1 \Rightarrow t=\frac{x-1}{1-2 x} \\
g(x)=\frac{x-1}{1-2 x}+1=\frac{x-1+1-2 x}{1-2 x} \\
g(x)=\frac{-x}{1-2 x} \\
\therefore \int g(x) d x=\int \frac{-x}{1-2 x} d x \\
=\frac{1}{2} \int \frac{1-2 x-1}{1-2 x} d x=\frac{1}{2}\left[\int 1 \cdot d x-\int \frac{1}{1-2 x} d x\right] \\
=\frac{1}{2}\left[x+\frac{1}{2} \log |2 x-1|\right]=\frac{x}{2}+\frac{1}{4} \log _e|2 x-1|+C .
\end{gathered}
$
Asked in: AP EAMCET 2018 (24 Apr Shift 1)
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