If \(\int \frac{\left(x^4+1\right)}{x\left(x^2+1\right)^2} d x=A \log |x|+\frac{B}{1+x^2}+c\), then \(\mathrm{A}-\mathrm{B}\) is (where c is the constant of integration)
Show Hint
Split the integrand into partial fractions, integrate, and compare with the given form to read A and B.
Step 1: Differentiate the given form
$\dfrac{d}{dx}\left[A\log|x|+\dfrac{B}{1+x^{2}}\right]=\dfrac{A}{x}-\dfrac{2Bx}{(1+x^{2})^{2}}$.
Step 2: Match with the integrand
$\dfrac{A(1+x^{2})^{2}-2Bx^{2}}{x(1+x^{2})^{2}}=\dfrac{x^{4}+1}{x(1+x^{2})^{2}}$.
Numerator: $Ax^{4}+(2A-2B)x^{2}+A=x^{4}+1$.
Step 3: Solve
$A=1$ and $2A-2B=0\Rightarrow B=1$, so $A-B=0$.
Final Answer:
A - B is 0.
\[ \boxed{\text{(A) }0} \]