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Question

Mathematics Question on integral

∫dxx(x2+1) equals∫\frac {dx}{x(x^2+1)} \ equals

A

log∣x∣−12log(x2+1)+Clog|x|-\frac 12log(x^2+1)+C

B

log∣x∣+12log(x2+1)+Clog|x|+\frac 12log(x^2+1)+C

C

−log∣x∣+12log(x2+1)+C-log|x|+\frac 12log(x^2+1)+C

D

12log∣x∣+log(x2+1)+C\frac 12log|x|+log(x^2+1)+C

Answer

log∣x∣−12log(x2+1)+Clog|x|-\frac 12log(x^2+1)+C

Explanation

Solution

Let 1x(x2+1)\frac {1}{x(x^2+1)} = Ax+Bx+Cx2+1\frac Ax+\frac {Bx+C}{x^2+1}

1=A(x2+1)+(Bx+C)x1 = 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

∴ 1x(x2+1)\frac {1}{x(x^2+1)} = 1x+−xx2+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).