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

Concept: U Substitution (standard form of inverse hyperbolic sine).

We want $\int \frac{1}{\sqrt{1+4x^2}} \, dx$.

Step 1: Factor the constant inside the square root to match the form $\frac{1}{\sqrt{1+u^2}}$.

Step 2: Let $u = 2x$, so $du = 2\,dx$, hence $dx = \frac{du}{2}$.

Step 3: Substitute and integrate using the standard result $\int \frac{1}{\sqrt{1+u^2}} \, du = \sinh^{-1} u + C$.

Step 4: Replace $u$ with $2x$.

Why This Approach Works

When you see $\sqrt{1 + 4x^2}$, your first thought should be: this looks like a Pythagorean identity. The expression $1 + (2x)^2$ under a square root is a dead giveaway for the identity $1 + \tan^2 \theta = \sec^2 \theta$. That’s the heart of it — we want to turn the square root into something clean like $\sec \theta$, which integrates nicely.

The substitution $2x = \tan \theta$ is the natural choice. It transforms the messy square root into a simple trigonometric function, and the $dx$ term will bring in a $\sec^2 \theta$ that cancels beautifully.

Step-by-Step Solution

1. Set up the substitution.

Let $2x = \tan \theta$. Then $x = \frac12 \tan \theta$, and differentiating gives:

2. Rewrite the square root.

The expression under the square root becomes:

So $\sqrt{1 + 4x^2} = \sqrt{\sec^2 \theta} = |\sec \theta|$. For the standard indefinite integral, we assume the domain where $\sec \theta > 0$ (typically $-\pi/2 < \theta < \pi/2$), so we can drop the absolute value:

3. Substitute everything into the integral.

The original integral is:

Substituting $dx = \frac12 \sec^2 \theta \, d\theta$ and $\sqrt{1+4x^2} = \sec \theta$:

4. Integrate $\sec \theta$.

The integral of $\sec \theta$ is a standard result:

So we have:

5. Convert back to $x$.

We know $\tan \theta = 2x$. To find $\sec \theta$, use the identity $\sec^2 \theta = 1 + \tan^2 \theta$:

Therefore:

6. Write the final answer.

Substituting back:

Method: Reduce to the standard form $\displaystyle\int\frac{dx}{\sqrt{x^2+a^2}}$

Integrands like $\dfrac{1}{\sqrt{1+4x^2}}$ are handled by factoring the coefficient of $x^2$ out of the square root to match a standard formula.

Steps

Step 1: Pull the coefficient out of the radical.

so the integrand becomes $\dfrac{1}{2}\cdot\dfrac{1}{\sqrt{x^2+(1/2)^2}}$.

Step 2: Apply the standard result.

with $a=\tfrac12$.

Step 3: Reassemble with the constant factor.

Carry the $\tfrac12$ through; the result is $\tfrac12\log\left|2x+\sqrt{1+4x^2}\right|+C$ (equivalent to the $a=\tfrac12$ form up to a constant).

The general move: whenever $x^2$ has a coefficient, factor it out to expose the underlying $\sqrt{x^2\pm a^2}$ standard form.

Common Mistakes

Mistake 1: Treating $\sqrt{1+4x^2}$ as if it were $\sqrt{1+x^2}$ and ignoring the coefficient.

Why it's wrong: the $4$ multiplying $x^2$ changes the effective $a$ and introduces a constant factor; skipping it gives a wrong argument. Correct approach: factor $\sqrt{1+4x^2}=2\sqrt{x^2+\tfrac14}$ so $a=\tfrac12$.

Mistake 2: Confusing the $\sqrt{x^2+a^2}$ (log) form with the $a^2-x^2$ (arcsine) form.

Why it's wrong: a plus under the root gives a logarithm/$\sinh^{-1}$, not $\sin^{-1}$. Correct approach: since it is $x^2+a^2$, use $\int\frac{dx}{\sqrt{x^2+a^2}}=\log|x+\sqrt{x^2+a^2}|+C$, giving $\tfrac12\log|2x+\sqrt{1+4x^2}|+C$.