Q.Integrate the following function: $\frac{x-1}{\sqrt{x^2-1}}$

Split the fraction so each piece is a standard form:

First term. Let $u = x^2-1$, so $du = 2x\,dx$:

Second term is the standard result $\int \frac{dx}{\sqrt{x^2-1}} = \log\left|x+\sqrt{x^2-1}\right|.$ …

The idea

The numerator $x-1$ is two pieces bundled together. The $x$ piece is (up to a factor of 2) exactly the derivative of $x^2-1$ sitting under the root — a textbook cue for $u$-substitution. The $-1$ piece leaves $\dfrac{1}{\sqrt{x^2-1}}$, a standard integral. So split first, then handle each part.

Step 1 — the first term by substitution

Let $u = x^2-1$, so $du = 2x\,dx$, i.e. $x\,dx = \tfrac12\,du$:

Check: $\dfrac{d}{dx}\sqrt{x^2-1} = \dfrac{x}{\sqrt{x^2-1}}.$ Correct.

Step 2 — the second term is standard

$\int \frac{dx}{\sqrt{x^2-1}} = \log\left|x+\sqrt{x^2-1}\right| + C.$ …

Method: Split a fraction into a "derivative-of-radical" part plus a standard form

For $\dfrac{x-1}{\sqrt{x^2-1}}$, separate the numerator so one piece matches the derivative of the radical (integrates to the radical itself) and the other is a standard reciprocal-radical form.

Steps

Step 1: Split the numerator.

Step 2: Integrate the first piece by substitution.

For $\dfrac{x}{\sqrt{x^2-1}}$, let $u=x^2-1$, $du=2x\,dx$, giving $\displaystyle\int\frac{x\,dx}{\sqrt{x^2-1}}=\sqrt{x^2-1}$.

Step 3: Integrate the second piece with a standard form. …

Common Mistakes

Mistake 1: Trying a single substitution for the whole $\dfrac{x-1}{\sqrt{x^2-1}}$.

Why it's wrong: the numerator mixes an $x$ (whose derivative-of-radical structure fits) with a constant $-1$ (which does not), so one substitution cannot handle both. Correct approach: split into $\dfrac{x}{\sqrt{x^2-1}}-\dfrac{1}{\sqrt{x^2-1}}$ and integrate each part.

Mistake 2: Using the wrong standard form for $\dfrac{1}{\sqrt{x^2-1}}$. …