Question

If $\int \frac{x^2+1}{x(x+1{)}^2}dx=A\ln |x|+\frac{B}{1+x}+C$, where C is the constant of integration, then $A+B$ $=$

• (A) 5
• (B) 4
• (C) 3
• (D) 2

Answer 1

The Correct Option is C

Explanation:
Given:$\int \frac{x^2+1}{x(x+1{)}^2}dx=A\ln |x|+\frac{B}{1+x}+C$$\int \frac{x^2+1}{x(x+1{)}^2}$$=\int \frac{(x+1{)}^2-2x}{x(x+1{)}^2}$$=\int (\frac{1}{x}-\frac{2}{(1+x{)}^2}dx)$$=\ln |x|-2\frac{(1+x{)}^{-1}}{-1}+C$$=\ln |x|+\frac{2}{1+x}+C$Comparing with the $RHS$,$A=1,B=2$$\therefore A+B=3$.Hence, the correct option is (C).