Q. \[ \int \frac{1}{x^2-1}\,dx \]

We want to compute the indefinite integral

\[
\int \frac{1}{x^2-1}\,dx.
\]

Step 1: Factor the denominator.

The expression \(x^2-1\) is a difference of squares:

\[
x^2-1=(x-1)(x+1).
\]

So the integrand becomes:

\[
\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}.
\]

Step 2: Use partial fraction decomposition.

We write

\[
\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.
\]

Multiply both sides by \((x-1)(x+1)\) to clear denominators:

\[
1=A(x+1)+B(x-1).
\]

Step 3: Solve for \(A\) and \(B\).

Expand the right-hand side:

\[
A(x+1)+B(x-1)=Ax+A+Bx-B.
\]

Group like terms:

\[
(A+B)x+(A-B).
\]

This must equal the constant \(1\), which is \(0\cdot x+1\). Therefore, match coefficients:

\[
A+B=0
\]
\[
A-B=1.
\]

Solve the system.

From \(A+B=0\), we get \(B=-A\).

Substitute into \(A-B=1\):

\[
A-(-A)=1
\]
\[
2A=1
\]
\[
A=\frac{1}{2}.
\]

Then

\[
B=-\frac{1}{2}.
\]

Step 4: Rewrite the integrand using \(A\) and \(B\).

\[
\frac{1}{x^2-1}=\frac{1/2}{x-1}-\frac{1/2}{x+1}
=\frac{1}{2}\cdot\frac{1}{x-1}-\frac{1}{2}\cdot\frac{1}{x+1}.
\]

Step 5: Integrate term-by-term.

Use the logarithm rule

\[
\int \frac{1}{x-a}\,dx=\ln|x-a|+C.
\]

So:

\[
\int \frac{1}{x^2-1}\,dx
=\int \left(\frac{1}{2}\cdot\frac{1}{x-1}-\frac{1}{2}\cdot\frac{1}{x+1}\right)\,dx.
\]

Integrate each part:

\[
=\frac{1}{2}\ln|x-1|-\frac{1}{2}\ln|x+1|+C.
\]

Step 6: Combine logarithms (optional).

Recall

\[
\ln|x-1|-\ln|x+1|=\ln\left|\frac{x-1}{x+1}\right|.
\]

Therefore the final answer is:

\[
\boxed{\int \frac{1}{x^2-1}\,dx=\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+C.}
\]