Definite Integration
Integration
Grade Class 12
Question:
If $f\left(\frac{1-x}{1+x}\right)=x$ and $g(x)=\int f(x)dx$ then<br>(A) $g(x)$ is continuous in domain<br>(B) $g(x)$ is discontinuous at two points in its domain<br>(C) $\lim_{x\to\infty} g'(x)=-1$<br>(D) $\int g(x)dx=-\frac{x^2}{2}+(2x+1)\ln\left(\frac{1+x}{e}\right)+C$
(A) $g(x)$ is continuous in domain
(B) $g(x)$ is discontinuous at two points in its domain
(C) $\lim_{x\to\infty} g'(x)=-1$
(D) $\int g(x)dx=-\frac{x^2}{2}+(2x+1)\ln\left(\frac{1+x}{e}\right)+C$
Step-by-Step Solution
Key Concept: Find f(x) by substituting t = (1-x)/(1+x), then integrate to find g(x) and check the properties.
Let $t = \frac{1-x}{1+x}$. Then $t+tx = 1-x \implies x(1+t) = 1-t \implies x = \frac{1-t}{1+t}$. Thus $f(t) = \frac{1-t}{1+t} = \frac{2-(1+t)}{1+t} = \frac{2}{1+t}-1$. So $f(x) = \frac{2}{1+x}-1$. Then $g(x) = \int (\frac{2}{1+x}-1) dx = 2\ln|1+x|-x+C$. $g(x)$ is defined for $x \neq -1$. It is continuous in its domain. $g'(x) = f(x) = \frac{2}{1+x}-1$. As $x \to \infty$, $g'(x) \to -1$.
Correct Answer: 1, 3