Q.Integrate the following function: $\sqrt{x^2+3x}$

Concept: U Substitution (completing the square inside the radical).

First, complete the square inside the square root:

Let $u = x + \frac{3}{2}$, so $du = dx$ and the integral becomes:

$\int \sqrt{u^2 - \left(\frac{3}{2}\right)^2} \, du$ …

The key to integrating expressions like $\sqrt{x^2+3x}$ is to recognize that the quadratic under the square root can be rewritten as a perfect square plus a constant. This is the technique of completing the square. Once we have something like $\sqrt{(x+a)^2 + b}$ or $\sqrt{(x+a)^2 - b}$, we can use a trigonometric substitution to eliminate the square root.

Why does this work? The identity $\sec^2\theta - 1 = \tan^2\theta$ is perfect for handling expressions of the form $\sqrt{u^2 - a^2}$, because substituting $u = a\sec\theta$ turns the square root into $a\tan\theta$, which is a simple trigonometric function. Similarly, $\sin^2$ and $\cos^2$ handle sums. Here, after completing the square, we get a difference of squares, so secant substitution is the right tool.

Let’s work through it step by step.

1. Complete the square. The expression inside the square root is $x^2 + 3x$. To complete the square, take half of the coefficient of $x$ (which is $\frac{3}{2}$), square it ($\frac{9}{4}$), and add and subtract it:

So the integral becomes:

2. Make a substitution to simplify. Let $u = x + \frac{3}{2}$, so $du = dx$. Then the integral is:

This is now in the standard form $\int \sqrt{u^2 - a^2} \, du$ with $a = \frac{3}{2}$.

3. Apply trigonometric substitution. For $\sqrt{u^2 - a^2}$, we use $u = a\sec\theta$, so $du = a\sec\theta\tan\theta \, d\theta$. Here $a = \frac{3}{2}$, so:

Then $\sqrt{u^2 - a^2} = \sqrt{\frac{9}{4}\sec^2\theta - \frac{9}{4}} = \frac{3}{2}\sqrt{\sec^2\theta - 1} = \frac{3}{2}\tan\theta$ (assuming $\tan\theta \geq 0$ for the principal branch; we’ll handle absolute values later).

The integral becomes:

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

Now we need to integrate $\sec^3\theta$ and $\sec\theta$. The integral of $\sec\theta$ is standard: $\int \sec\theta \, d\theta = \log|\sec\theta + \tan\theta| + C$.

For $\sec^3\theta$, we use integration by parts or a known reduction formula. Let’s do it quickly:

(This can be derived by writing $\int \sec^3\theta \, d\theta = \int \sec\theta \cdot \sec^2\theta \, d\theta$ and integrating by parts with $u = \sec\theta$, $dv = \sec^2\theta \, d\theta$.)

So:

Simplify the log terms: $\frac{1}{2}\log|\cdots| - \log|\cdots| = -\frac{1}{2}\log|\sec\theta + \tan\theta|$.

Thus:

5. Back-substitute to $u$ and then to $x$. We have $u = \frac{3}{2}\sec\theta$, so $\sec\theta = \frac{2u}{3}$. Also, $\tan\theta = \sqrt{\sec^2\theta - 1} = \sqrt{\frac{4u^2}{9} - 1} = \frac{2}{3}\sqrt{u^2 - \frac{9}{4}} = \frac{2}{3}\sqrt{u^2 - a^2}$. But note: $\sqrt{u^2 - a^2}$ is exactly the original square root we had! So $\tan\theta = \frac{2}{3}\sqrt{u^2 - \frac{9}{4}}$. …

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: Forgetting the constant term while completing the square of $x^{2}+3x$.

Why it's wrong: $x^{2}+3x=(x+\tfrac32)^{2}-\tfrac94$; the $-\tfrac94$ is easy to omit when the original has no constant. Correct approach: add and subtract $(3/2)^{2}=\tfrac94$.

Mistake 2: Using the arcsine template. …