Question

If $\left(\frac{1-x}{1+x}\right)=f\left(x\right)$ and $g\left(x\right)=\int f\left(x\right)dx$ then

• A. $g\left(x\right)$ is continuous in domain
• B. $g\left(x\right)$ is discontinuous at two points in its domain
• C. $\underset{x\to \infty }{lim}g\left(x\right)=-1$
• D. $\int g\left(x\right)dx=-\frac{x^2}{2}+(2x+1)\lambda n\left(\frac{1+x}{e}\right)+C$

Answer 1

The correct option is A $g\left(x\right)$ is continuous in domain

$f\left(x\right)=\frac{1-x}{1+x}$ and $g\left(x\right)=\int f\left(x\right)dx$

$\Rightarrow g\left(x\right)=\int f\left(x\right)dx$

$\Rightarrow g\left(x\right)=\int \frac{1-x}{1+x}dx$

$\Rightarrow g\left(x\right)=\int \frac{1}{1+x}dx-\int \frac{x}{1+x}dx$

$\Rightarrow g\left(x\right)=\int \frac{1}{1+x}dx-\int \frac{x+1-1}{1+x}dx$

$\Rightarrow g\left(x\right)=\int \frac{1}{1+x}dx-\int \frac{x+1}{1+x}dx+\int \frac{1}{1+x}dx$

$\Rightarrow g\left(x\right)=2\int \frac{1}{1+x}dx-\int 1dx$

$\Rightarrow g\left(x\right)=2\log |1+x|-x$

Domain of g(x) does not include $1$

Hence g(x) is continuous in the domain