$\displaystyle
\int\frac{dx}{1-x^2}
$
I am supposed to use partial fractions to solve this. I am pretty sure you could use a trig substitution here but how would you use partial fractions with this integral?
$\displaystyle \frac{1}{1-x^2}=\frac{1}{(1+x)(1-x)}$
$\displaystyle \frac{1}{(1+x)(1-x)}=\frac{A}{1+x}+\frac{B}{1-x}$
$\displaystyle \frac{1}{(1+x)(1-x)}= \frac{A(1-x)+B(1+x)}{(1+x)(1-x)}$
$\displaystyle \frac{1}{(1+x)(1-x)}=\frac{(A+B)+(A-B)x}{(1+x)(1-x)}$
Now, in the numerator of both sides, equate the coefficients of constant terms, and x-terms,
A + B =1
A - B =0
Solving we get, A = B = 1/2
so, $\displaystyle \int {\frac{1}{1-x^2}}~dx= \int {\left(\frac{1/2}{1+x}+\frac{1/2}{1-x}\right)}~dx$
$\displaystyle =\frac{1}{2}\int{\left( \frac{1}{1+x}+\frac{1}{1-x}\right)}~dx$