d x x (x^ 2 +1 ) equals

Question

$\int \frac{d x}{x\left(x^{2}+1\right)}$ equals

✓

Answer

Let $\frac{1}{x\left(x^{2}+1\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$
$1 = A(X^2 + 1) + (Bx + C)x = Ax^2 + A + Bx^2 + Cx = (A + B)x^2+Cx + A$
Equating the coefficients of $x^2, x$ and constant term, we get,
$A + B = 0$
$C = 0$
$A = 1$
On solving these equations, we get,
$A = 1, B = -1$ and $C = 0$
Therefore, $\frac{1}{x\left(x^{2}+1\right)}=\frac{1}{x}+\frac{-x}{x^{2}+1}$
$\int \frac{1}{x\left(x^{2}+1\right)}=\int\left\{\frac{1}{x}+\frac{-x}{x^{2}+1}\right\} d x$
$= \log |x|-\frac{1}{2} \log \left|x^{2}+1\right|+C$

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