Q.Evaluate: $\int \sqrt{5-2x+x^2}\,dx$

Concept: Integration by completing the square, then the standard integral for $\sqrt{x^2+a^2}$.

Complete the square inside the radical:

So the integral is $\displaystyle\int \sqrt{(x-1)^2+2^2}\,dx$, which matches the standard form

with $u=x-1$ and $a=2$. Substituting back (and using $(x-1)^2+4=x^2-2x+5$): …

The plan

An integral of the form $\int\sqrt{\text{quadratic}}\,dx$ is handled in two moves: first complete the square so the quadratic becomes $(x-h)^2+k^2$, then use the memorised standard integral for $\sqrt{u^2+a^2}$.

Step 1 — Complete the square

Re-order the quadratic as $x^2-2x+5$ and complete the square:

So

Step 2 — Substitute

Put $u=x-1$, $du=dx$:

Step 3 — Apply the standard integral

The CBSE standard result is

With $a=2$ (so $a^2=4$, $\tfrac{a^2}{2}=2$):

$\int \sqrt{u^2+4}\,du = \frac{u}{2}\sqrt{u^2+4} + 2\log\left|u+\sqrt{u^2+4}\right| + C.$ …

Method: Integrating $\sqrt{\text{quadratic}}$ by completing the square

Use this for $\int\sqrt{\text{quadratic}}\,dx$ where the quadratic has a positive $x^2$. Complete the square to $u^2+a^2$, then apply the standard $\int\sqrt{u^2+a^2}\,du$ formula.

Steps

Step 1: Complete the square.

so set $u=x-1$, $a^2=4$.

Step 2: Recall the standard integral.

Step 3: Substitute $u=x-1$, $a^2=4$. …

Common Mistakes

Mistake 1: Using the wrong standard template.

Why it's wrong: $\int\sqrt{u^2+a^2}\,du$ has a $\log$/arcsinh second term, while $\int\sqrt{a^2-u^2}\,du$ has an arcsine — mixing them is a common slip. Correct approach: match the sign inside the root ($u^2+a^2$ here).

Mistake 2: Forgetting the $\tfrac{a^2}{2}$ coefficient. …