Partial Fractions — Question 7

Question 7

Use partial fractions to evaluate exactly: $$\int_0^{1/2}\frac{dx}{1-x^2}.$$ Why is this integral proper on $[0,1/2]$?

Question 7 – Solution

Step 1: Factor and set up partial fractions. $$1-x^2=(1-x)(1+x),$$ $$\frac1{1-x^2}=\frac{A}{1-x}+\frac{B}{1+x}.$$ Step 2: Clear denominators and find $A$ and $B$. $$1=A(1+x)+B(1-x).$$ $$\begin{align*} x=1:&\quad 1=2A \Longrightarrow A=\frac12,\\ x=-1:&\quad 1=2B \Longrightarrow B=\frac12. \end{align*}$$ Step 3: Integrate and evaluate both bounds. $$\begin{align*} I&=\frac12\int_0^{1/2}\left(\frac1{1-x}+\frac1{1+x}\right)dx\\ &=\frac12\left[-\ln(1-x)+\ln(1+x)\right]_0^{1/2}\\ &=\frac12\left[-\ln\frac12+\ln\frac32-(-\ln1+\ln1)\right]\\ &=\frac12\ln\left(\frac{3/2}{1/2}\right)=\frac12\ln3. \end{align*}$$ Step 4: Check whether the integral is proper. The singularities $x=\pm1$ are outside $[0,1/2]$, so the integrand is continuous on the whole interval and no improper limit is needed. $$\boxed{\displaystyle \int_0^{1/2}\frac{dx}{1-x^2}=\frac12\ln3}.$$