Partial Fractions — Question 4

Question 4 – Solution

Step 1: Set up the decomposition. Use a linear numerator over the irreducible quadratic: $$\frac{x^2+1}{x(x^2+4)}=\frac{A}{x}+\frac{Bx+C}{x^2+4}.$$ Step 2: Clear denominators and collect like terms. $$\begin{align*} x^2+1&=A(x^2+4)+x(Bx+C)\\ &=(A+B)x^2+Cx+4A. \end{align*}$$ Step 3: Match coefficients. $$A+B=1,\qquad C=0,\qquad 4A=1.$$ Therefore $A=1/4$, $B=1-1/4=3/4$, and $C=0$.

Step 4: Integrate. For the second term, let $u=x^2+4$; then $du=2x\,dx$. $$\begin{align*} I&=\frac14\int\frac{dx}{x} +\frac34\int\frac{x}{x^2+4}\,dx\\ &=\frac14\ln|x|+\frac34\left(\frac12\int\frac{du}{u}\right)\\ &=\frac14\ln|x|+\frac38\ln(x^2+4)+C. \end{align*}$$ Thus, $$\boxed{\displaystyle I=\frac14\ln|x|+\frac38\ln(x^2+4)+C}.$$