Q.Integrate the function $\sqrt{1+\frac{x^2}{9}}$

The key idea is to rewrite the integrand in a form that matches the standard formula for $\int \sqrt{a^2 + x^2} \, dx$, using completing the square (here, the square is already complete).

First, simplify the square root:

Now integrate:

Use the standard formula $\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \sinh^{-1}\frac{x}{a} + C$ (or the equivalent logarithmic form). Here $a = 3$, so: …

The expression $\sqrt{1+\frac{x^2}{9}}$ looks like it came straight from a right triangle. When you see $1 + (\text{something})^2$ under a square root, your mind should immediately go to one of two places: either a trigonometric substitution (like $x = a\tan\theta$) or a hyperbolic substitution (like $x = a\sinh t$). Both work; the choice is a matter of taste.

The key insight: the constant $1$ and the fraction $\frac{x^2}{9}$ are not in the simplest form for substitution. Factor out the $\frac{1}{9}$ first.

1. Simplify the integrand algebraically

So the integral becomes:

Now we have a clean $\sqrt{x^2 + a^2}$ form with $a=3$.

2. Choose the substitution

   For $\sqrt{x^2 + a^2}$, the standard trigonometric substitution is $x = a\tan\theta$. Why? Because $1+\tan^2\theta = \sec^2\theta$, which turns the square root into something simple.

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

   Watch out

   A common mistake: forgetting to also change $dx$ when substituting. The $dx$ is not $d\theta$ — you must multiply by the derivative.

3. Rewrite the integrand in $\theta$

For the principal range of $\theta = \tan^{-1}(x/3)$, we have $\theta \in (-\pi/2, \pi/2)$, where $\sec\theta > 0$. So we can drop the absolute value: $\sqrt{x^2+9} = 3\sec\theta$.

Therefore:

4. Integrate $\sec^3\theta$

   This is a classic integral. The trick: write $\sec^3\theta = \sec\theta \cdot \sec^2\theta$ and integrate by parts.

   Let $u = \sec\theta$, $dv = \sec^2\theta\,d\theta$. Then $du = \sec\theta\tan\theta\,d\theta$, $v = \tan\theta$.

Now use $\tan^2\theta = \sec^2\theta - 1$:

Bring the $\int \sec^3\theta$ term to the left:

And $\int \sec\theta\,d\theta = \log|\sec\theta + \tan\theta| + C$.

So:

5. Substitute back to $x$

   We have $I = 3\int \sec^3\theta\,d\theta$, so:

Now recall: $x = 3\tan\theta$, so $\tan\theta = \frac{x}{3}$. …

Method: Factor out the constant, then use a standard $\sqrt{t^{2}+a^{2}}$ formula

When the quadratic under the root has a fractional or non-unit coefficient, e.g. $\sqrt{1+\frac{x^{2}}{9}}$, pull the constant out of the root to expose a clean $\sqrt{x^{2}+a^{2}}$ form.

Steps

Step 1: Combine into a single fraction under the root.

Step 2: Identify $a^{2}$.

Now it is $\dfrac13\sqrt{x^{2}+a^{2}}$ with $a^{2}=9$, $a=3$.

Step 3: Apply the standard integral. …

Common Mistakes

Mistake 1: Pulling the constant out as $\dfrac19$ instead of $\dfrac13$.

Why it's wrong: $\sqrt{\dfrac{9+x^{2}}{9}}=\dfrac{1}{\sqrt9}\sqrt{9+x^{2}}=\dfrac13\sqrt{9+x^{2}}$ — the root of $9$ is $3$. Correct approach: take the square root of the denominator.

Mistake 2: Using $a^{2}=3$. …