Q.Integrate the following function: $\sqrt{x^2+4x-5}$

The key idea is Integration By Completing the Square — rewriting the quadratic inside the square root to use standard trigonometric substitution forms.

First, complete the square:

So the integral becomes $\int \sqrt{(x+2)^2 - 9} \, dx$.

Let $u = x+2$, then $du = dx$, and we have $\int \sqrt{u^2 - 9} \, du$. This matches the standard form $\int \sqrt{u^2 - a^2} \, du$ with $a = 3$.

Using the known formula:

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

Why Completing the Square Works

When you see a quadratic inside a square root, your first instinct might be to try a $u$-substitution. But $x^2+4x-5$ isn't a perfect square — it has a linear term that blocks a clean substitution. Completing the square rewrites it as $(x+2)^2 - 9$, which is a difference of squares. That form screams for a trigonometric substitution, because expressions like $\sqrt{u^2 - a^2}$ are tailor-made for secant or hyperbolic cosine substitutions.

The core idea: turn the messy quadratic into a recognizable Pythagorean form, then let trigonometry handle the rest.

Step-by-Step Solution

1. Complete the square inside the radical.

Take $x^2+4x-5$. Half of 4 is 2, square it to get 4. Add and subtract 4:

So the integral becomes:

2. Choose a substitution that eliminates the square root.

We have $\sqrt{u^2 - a^2}$ with $u = x+2$ and $a = 3$. The standard trick: set $u = a\sec\theta$, because $\sec^2\theta - 1 = \tan^2\theta$.

Let $x+2 = 3\sec\theta$. Then $dx = 3\sec\theta\tan\theta \, d\theta$.

Now substitute:

For the indefinite integral, we assume the principal branch where $\theta \in [0, \pi/2) \cup (\pi/2, \pi]$, so $\tan\theta \ge 0$ when $\sec\theta \ge 1$ (which matches $x+2 \ge 3$). We'll take $\tan\theta \ge 0$ and drop the absolute value, keeping in mind the domain.

3. Rewrite the integral in terms of $\theta$.

4. Simplify $\tan^2\theta$ using the Pythagorean identity.

Recall $\tan^2\theta = \sec^2\theta - 1$. So:

Now we need two standard integrals: $\int \sec^3\theta \, d\theta$ and $\int \sec\theta \, d\theta$.

5. Apply the formulas.

Subtract: …

Method: Complete the square, then use a standard $\sqrt{\ }$ formula

To integrate $\sqrt{\text{quadratic}}$, rewrite the quadratic as $(x+p)^{2}\pm a^{2}$ or $a^{2}-(x+p)^{2}$ by completing the square, substitute $t=x+p$, and quote the matching standard integral.

Steps

Step 1: Complete the square on the quadratic under the root, so it becomes $(x+p)^{2}+k$ for some constant $k$.

Step 2: Substitute $t=x+p$ (so $dt=dx$); the integral becomes $\int\sqrt{t^{2}\pm a^{2}}\,dt$ or $\int\sqrt{a^{2}-t^{2}}\,dt$.

Step 3: Apply the correct standard formula.

$\int\sqrt{t^{2}-a^{2}}\,dt=\frac{t}{2}\sqrt{t^{2}-a^{2}}-\frac{a^{2}}{2}\log\big|t+\sqrt{t^{2}-a^{2}}\big|+C,$ …

Common Mistakes

Mistake 1: Completing the square as $(x+2)^{2}-5\Rightarrow (x+2)^{2}+ \dots$ sign slip.

Why it's wrong: $x^{2}+4x-5=(x+2)^{2}-9$; the constant is $-9$, a $t^{2}-a^{2}$ form. Correct approach: $(x+2)^{2}-4-5=(x+2)^{2}-9$, so $a^{2}=9$.

Mistake 2: Using $\sin^{-1}$ instead of $\log$. …