Question

Assertion :Consider the function $F\left(x\right)=\int \frac{x}{(x-1)(x^2+1)}dx$
STATEMENT-1 : $F\left(x\right)$ is discontinuous at $x=1$ Reason: STATEMENT-2 : Integrand of $F\left(x\right)$ is discontinuous at $x=1$

• A. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
• B. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
• C. STATEMENT-1 is True, STATEMENT-2 is False
• D. STATEMENT-1 is False, STATEMENT-2 is True

Answer 1

The correct option is B STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

Given, $f\left(x\right)=\int \frac{x}{(x-1)(x^2+1)}dx\frac{x}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}A=1,B=-1,C=1f\left(x\right)=\int \frac{1}{x-1}dx-\int \frac{x-1}{x^2+1}dxf\left(x\right)=\ln (x-1)-\int \frac{x}{x^2+1}dx+\int \frac{dx}{x^2+1}f\left(x\right)=\ln (x-1)-\frac{\ln x^2+1}{2}+{\tan }^{-1}x+C$

$\ln (x-1)$ is not continous at $x=1$