Q.Evaluate the definite integral: $\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}$ equals (A) $\frac{\pi}{3}$ (B) $\frac{2\pi}{3}$ (C) $\frac{\pi}{6}$ (D) $\frac{\pi}{12}$

Concept: Definite Integral of $\frac{1}{1+x^2}$ — the antiderivative is $\tan^{-1} x$.

Step 1: Recall the standard result

$\int \frac{dx}{1+x^2} = \tan^{-1} x + C$.

Step 2: Apply the limits

$\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} = \left[ \tan^{-1} x \right]_{1}^{\sqrt{3}} = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1)$.

Step 3: Evaluate the arctangents …

The key here is recognising that $\frac{1}{1+x^2}$ is the derivative of $\arctan x$. This is one of the most fundamental inverse trigonometric integrals — it appears constantly in calculus. The definite integral from $a$ to $b$ of $\frac{dx}{1+x^2}$ simply gives the difference of the arctangent values at the limits.

Let’s walk through it.

1. Recall the antiderivative. We know that

This is a direct consequence of the fact that $\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$.

2. Apply the limits of integration. Using the Fundamental Theorem of Calculus:

3. Evaluate each arctangent.

   • $\arctan(1)$ is the angle whose tangent is $1$. That angle is $\frac{\pi}{4}$ (since $\tan \frac{\pi}{4} = 1$).
   • $\arctan(\sqrt{3})$ is the angle whose tangent is $\sqrt{3}$. That angle is $\frac{\pi}{3}$ (since $\tan \frac{\pi}{3} = \sqrt{3}$).

   So we have:

4. Subtract the fractions. …

Method: Recognise the standard inverse-tangent integral

The form $\displaystyle\int \frac{dx}{1+x^2}$ is a memorised standard result: it is exactly the derivative rule for $\tan^{-1}x$ read backwards.

Steps

Step 1: Identify the form.

Whenever the integrand is $\dfrac{1}{1+x^2}$ (denominator $1+x^2$, numerator constant), reach straight for the arctangent antiderivative.

Step 2: Write the antiderivative.

Step 3: Apply the limits and evaluate the standard angles. …

Common Mistakes

Mistake 1: Confusing $\tan^{-1}\sqrt3$ with $\tfrac{\pi}{6}$.

Why it's wrong: $\tan\tfrac{\pi}{6}=\tfrac{1}{\sqrt3}$, whereas $\tan\tfrac{\pi}{3}=\sqrt3$, so $\tan^{-1}\sqrt3=\tfrac{\pi}{3}$. Correct approach: match the tangent value to the correct standard angle.

Mistake 2: Subtracting the fractions incorrectly. …