Definite Integration
Function substitution
Grade Class 12
Question:
If $f\left(\frac{1-x}{1+x}\right)=x$ and $g(x)=\int f(x)dx$ then
(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: First, find the expression for f(x) by substituting t = (1-x)/(1+x), which gives x = (1-t)/(1+t). Then f(t) = (1-t)/(1+t). Thus f(x) = (1-x)/(1+x) = -1 + 2/(1+x). Integrate f(x) to find g(x) = -x + 2ln|1+x| + C. Analyze the properties of g(x) and its integral.
Let $t = \frac{1-x}{1+x}$. Then $t+tx = 1-x \implies x(1+t) = 1-t \implies x = \frac{1-t}{1+t}$. So $f(t) = \frac{1-t}{1+t} = \frac{2-(1+t)}{1+t} = \frac{2}{1+t}-1$. Thus $f(x) = \frac{2}{1+x}-1$. Then $g(x) = \int (\frac{2}{1+x}-1) dx = 2\ln|1+x|-x+C$. The domain of $g(x)$ is $x \neq -1$. $g(x)$ is continuous in its domain. $g'(x) = f(x) = \frac{2}{1+x}-1$. As $x \to \infty$, $g'(x) \to -1$. $\int g(x)dx = \int (2\ln(1+x)-x+C)dx = 2((1+x)\ln(1+x)-(1+x)) - x^2/2 + Cx + C_1 = 2(1+x)\ln(1+x) - 2x - x^2/2 + Cx + C_1$.
Correct Answer: A,C