Q.Evaluate the definite integral: $\int_0^1 \frac{dx}{1+x^2}$

The key idea is that $\frac{1}{1+x^2}$ is the derivative of $\tan^{-1} x$, so we can integrate directly without substitution.

Step 1: Recall the standard integral:

Step 2: Apply the limits of integration: …

The key here is recognizing that $\frac{1}{1+x^2}$ is the derivative of $\arctan x$. This is one of the most fundamental inverse trigonometric integrals, and it appears often in calculus. The integral is already in its simplest form — no substitution is needed because the antiderivative is direct.

Let’s work through it step by step.

1. Identify the antiderivative The integral $\int \frac{dx}{1+x^2}$ is a standard result:

This comes from the fact that $\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$.

2. Apply the limits of integration We evaluate the definite integral from $0$ to $1$:

3. Evaluate the arctan values
   • $\arctan(1) = \frac{\pi}{4}$ (since $\tan(\pi/4) = 1$)
   • $\arctan(0) = 0$ (since $\tan(0) = 0$) …

Method: Definite integral giving an inverse tangent

Recognise $\int\dfrac{dx}{1+x^{2}}=\tan^{-1}x$, then evaluate across the limits.

Steps

Step 1: Match the standard form. $\dfrac{1}{1+x^{2}}$ integrates to $\tan^{-1}x$.

Step 2: Bracket it.

$\int_{a}^{b}\frac{dx}{1+x^{2}}=\big[\tan^{-1}x\big]_{a}^{b}.$ …

Common Mistakes

Mistake 1: Confusing $\dfrac{1}{1+x^{2}}$ with $\dfrac{1}{\sqrt{1-x^{2}}}$ (arcsine).

Why it's wrong: no square root here, so the antiderivative is $\tan^{-1}x$, not $\sin^{-1}x$. Correct approach: match $1+x^{2}\Rightarrow\tan^{-1}x$.

Mistake 2: Using $\tan^{-1}1=45$ (degrees) in a calculus answer. …