Question

integrals

Answer 1

\[\int\frac{dx}{a^2-x^2}=\int\frac{dx}{(a-x)(a+x)}\]
\[\begin{align*}\frac{1}{(a-x)(a+x)}&=\frac{A}{a-x}+\frac{B}{a+x}\\
1&=A(a+x)+B(a-x)\\
1&=(A-B)x+(A+B)a
\end{align*}\]
giving the system,
\[\begin{cases}A-B=0\\A+B=\dfrac{1}{a}\end{cases}~~\Rightarrow~~A=B=\frac{1}{2a}\]
So,
\[\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\left[\int\frac{dx}{a-x}+\int\frac{dx}{a+x}\right]=\cdots\]