Q.Integrate the following function: $\sqrt{x^2+4x+1}$

The key idea is to complete the square inside the square root, then use a trigonometric substitution.

First, complete the square:

So the integral becomes $\int \sqrt{(x+2)^2 - 3} \, dx$.

Let $u = x+2$, so $du = dx$. The integral is now $\int \sqrt{u^2 - 3} \, du$.

This is a standard form: $\int \sqrt{u^2 - a^2} \, du$ with $a = \sqrt{3}$. The formula is:

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

Why This Approach Works

When you see a quadratic inside a square root, your first instinct should be to complete the square. That turns the expression into something like $\sqrt{(x+h)^2 \pm k}$, which then suggests a trigonometric or hyperbolic substitution. Here, after completing the square, we get $\sqrt{(x+2)^2 - 3}$. That’s a perfect match for the secant substitution: $\sqrt{u^2 - a^2}$ with $u = x+2$ and $a = \sqrt{3}$. The secant substitution works because $\sec^2\theta - 1 = \tan^2\theta$, which lets the square root simplify cleanly.

Let’s walk through it.

1. Complete the square inside the radical. $x^2+4x+1 = (x^2+4x+4) - 3 = (x+2)^2 - 3$. So the integral becomes

2. Set up the substitution. Let $u = x+2$, so $du = dx$. Then

This is of the form $\sqrt{u^2 - a^2}$ with $a = \sqrt{3}$.

3. Use the secant substitution. For $\sqrt{u^2 - a^2}$, the standard trick is $u = a \sec\theta$, which gives $du = a \sec\theta \tan\theta \, d\theta$. Here $a = \sqrt{3}$, so let $u = \sqrt{3} \sec\theta$. Then $du = \sqrt{3} \sec\theta \tan\theta \, d\theta$. Now

For the domain where the original integrand is defined (we’ll assume $u \ge \sqrt{3}$ so $\theta \in [0, \pi/2)$), $\tan\theta \ge 0$, so we can drop the absolute value:

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

5. Simplify the trigonometric integral. Use $\tan^2\theta = \sec^2\theta - 1$:

Now we need two standard integrals:

• $\int \sec\theta \, d\theta = \log|\sec\theta + \tan\theta| + C$.
• $\int \sec^3\theta \, d\theta$ is a classic. Its derivation uses integration by parts:
  ›Proof

  Let $I = \int \sec^3\theta \, d\theta$. Write $\sec^3\theta = \sec\theta \cdot \sec^2\theta$. Integrate by parts: let $dv = \sec^2\theta \, d\theta$, $u = \sec\theta$. Then $v = \tan\theta$, $du = \sec\theta \tan\theta \, d\theta$.

  $I = \sec\theta \tan\theta - \int \tan\theta \cdot \sec\theta \tan\theta \, d\theta = \sec\theta \tan\theta - \int \sec\theta \tan^2\theta \, d\theta.$

  Replace $\tan^2\theta = \sec^2\theta - 1$:

  $I = \sec\theta \tan\theta - \int \sec\theta (\sec^2\theta - 1) \, d\theta = \sec\theta \tan\theta - \int \sec^3\theta \, d\theta + \int \sec\theta \, d\theta.$

  So $I = \sec\theta \tan\theta - I + \log|\sec\theta + \tan\theta|$, giving $2I = \sec\theta \tan\theta + \log|\sec\theta + \tan\theta|$, hence

  $\int \sec^3\theta \, d\theta = \frac{1}{2} \sec\theta \tan\theta + \frac{1}{2} \log|\sec\theta + \tan\theta| + C.$

Using this,

Simplify:

6. Back-substitute to $u$ (and then $x$). Recall $u = \sqrt{3} \sec\theta$, so $\sec\theta = \frac{u}{\sqrt{3}}$. Also $\tan\theta = \sqrt{\sec^2\theta - 1} = \sqrt{\frac{u^2}{3} - 1} = \frac{\sqrt{u^2 - 3}}{\sqrt{3}}$. …

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: Writing $x^{2}+4x+1=(x+2)^{2}+1$.

Why it's wrong: $(x+2)^{2}=x^{2}+4x+4$, so $x^{2}+4x+1=(x+2)^{2}-3$ — a minus, not plus. Correct approach: subtract the extra $3$.

Mistake 2: Using the $\sqrt{t^{2}+a^{2}}$ formula instead of $\sqrt{t^{2}-a^{2}}$. …