Question

\(∫\frac {dx}{x(x^2+1)} \ equals \)

• A. \(log|x|-\frac 12log(x^2+1)+C\)
• B. \(log|x|+\frac 12log(x^2+1)+C\)
• C. \(-log|x|+\frac 12log(x^2+1)+C\)
• D. \(\frac 12log|x|+log(x^2+1)+C\)

Answer 1

The Correct Option is A

Let \(\frac {1}{x(x^2+1)}\) = \(\frac Ax+\frac {Bx+C}{x^2+1}\)

\(1 = A(x^2+1)+(Bx+C)x\)

Equating the coefficients of x², x, and constant term, we obtain
A + B = 0
C = 0
A = 1
On solving these equations, we obtain
A = 1, B = −1, and C = 0

∴ \(\frac {1}{x(x^2+1)}\) = \(\frac 1x+\frac {-x}{x^2+1}\)

⇒ \(∫\)\(\frac {1}{x(x^2+1)} dx\) = \(∫\)\(\frac 1x-\frac {x}{x^2+1}dx\)

                               = \(log|x|-\frac 12log|x^2+1|+C\)

Hence, the correct Answer is (A).