Q.Evaluate $\int_0^1 \dfrac{\tan^{-1} x}{1 + x^2}\, dx$

The key idea is U Substitution: the derivative of $\tan^{-1} x$ is $\frac{1}{1+x^2}$, which appears in the integrand.

Let $u = \tan^{-1} x$. Then $du = \frac{1}{1+x^2}\, dx$.

When $x = 0$, $u = 0$; when $x = 1$, $u = \frac{\pi}{4}$.

The integral becomes: …

The key insight here is noticing the relationship between the numerator and denominator. The denominator $1 + x^2$ is exactly the derivative of $\tan^{-1} x$. This is a classic setup for a substitution — whenever you see a function and its derivative multiplied together (or in a ratio like this), substitution is the way to go.

Let’s walk through it.

1. Identify the substitution.

   Let $u = \tan^{-1} x$. Then $du = \frac{1}{1 + x^2} dx$. This is perfect because the entire integrand $\frac{\tan^{-1} x}{1 + x^2} dx$ becomes $u \, du$.

2. Change the limits of integration.

   When $x = 0$, $u = \tan^{-1} 0 = 0$.

   When $x = 1$, $u = \tan^{-1} 1 = \frac{\pi}{4}$.

   So the integral in $u$ runs from $0$ to $\frac{\pi}{4}$.

3. Rewrite and integrate.

   The original integral becomes:

This is a simple power rule integral:

4. Evaluate. …

Method: Definite Substitution Recognising a Function and Its Derivative

Use this when the integrand contains a function together with its own derivative, e.g. $\tan^{-1}x$ over $1+x^2$: substitute the function and convert the limits.

Steps

Step 1: Substitute the function whose derivative is present.

Note $\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$, which is exactly the $\dfrac{dx}{1+x^2}$ in the integrand. Set $u=\tan^{-1}x$, $du=\dfrac{dx}{1+x^2}$.

Step 2: Change the limits. …

Common Mistakes

Mistake 1: Missing that $\dfrac{1}{1+x^2}$ is the derivative of $\tan^{-1}x$.

Why it's wrong: without spotting this the substitution $u=\tan^{-1}x$ is never made. Correct approach: recognise the function-and-derivative pairing.

Mistake 2: Not converting the limits to $u$.

Why it's wrong: the new bounds are $0$ and $\frac\pi4$, not $0$ and $1$. Correct approach: apply $u=\tan^{-1}x$ to each limit. …