# Integral of 1/(1+x²): Formula, Steps and Examples

> The integral of 1/(1+x²) is arctan x + C. See why with a tan substitution, the 1/(a²+x²) version, a graph and worked examples.

Source: https://practicecalculus.com/integrals/1-over-1-plus-x-squared
Updated: 2026-11-06

$$ \int \frac{1}{1+x^2}\,dx = \arctan x + C $$

## At a glance

- **Antiderivative:** $\arctan x + C$
- **General form:** $\displaystyle\int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\frac{x}{a} + C$
- **Method:** Substitution $x = \tan\theta$, or recognize the arctan derivative
- **From 0 to 1:** $\dfrac{\pi}{4}$
- **Over the whole line:** $\pi$

## Key takeaways

- The integral of $\frac{1}{1+x^2}$ is $\arctan x + C$, because the derivative of $\arctan x$ is $\frac{1}{1+x^2}$.
- For a constant $a > 0$, $\int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\frac{x}{a} + C$. Do not forget the $\frac{1}{a}$ in front.
- If there is an $x$ in the numerator, as in $\frac{x}{1+x^2}$, the answer is a log, not an arctan.
- Quadratics like $x^2 + 2x + 5$ become arctan integrals after you complete the square.

The **integral of $\dfrac{1}{1+x^2}$ is $\arctan x + C$**. It is one of the few integrals you should know on sight, because it turns a plain rational function into an angle.

$$\int \frac{1}{1+x^2}\,dx = \arctan x + C$$

The more general version, for any constant $a > 0$, is

$$\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C$$

## Graph of 1/(1+x²) and its antiderivative

```graph
title: 'y = 1/(1+x²) with the area from 0 to 1 shaded'
caption: 'The shaded area equals $\arctan 1 - \arctan 0 = \frac{\pi}{4} \approx 0.785$. The dashed curve is the antiderivative $\arctan x$, which levels off at $\pm\frac{\pi}{2}$.'
x: [-5, 5]
y: [-2, 2]
functions:
  - expr: 1/(1+x^2)
    label: y = 1/(1+x²)
  - expr: arctan(x)
    label: y = arctan x (antiderivative)
    dashed: true
shade:
  expr: 1/(1+x^2)
  from: 0
  to: 1
points:
  - [0, 1, '(0, 1)']
```

The curve $\frac{1}{1+x^2}$ is positive everywhere, so its antiderivative is always increasing. It also flattens out fast, which is why $\arctan x$ approaches the horizontal lines $y = \pm\frac{\pi}{2}$.

## How to integrate 1/(1+x²) step by step

You can simply recognize the [derivative of arctan x](/derivatives/arctan-x). If you want to see where the answer comes from, use the substitution $x = \tan\theta$.

1. Let $x = \tan\theta$, so $dx = \sec^2\theta\,d\theta$.
2. Use the identity $1 + \tan^2\theta = \sec^2\theta$:
$$\int \frac{\sec^2\theta}{1+\tan^2\theta}\,d\theta = \int \frac{\sec^2\theta}{\sec^2\theta}\,d\theta = \int 1\,d\theta = \theta + C$$
3. Go back to $x$. Since $x = \tan\theta$, we have $\theta = \arctan x$.

For $\frac{1}{a^2+x^2}$, factor out $a^2$ and substitute $u = \frac{x}{a}$, $dx = a\,du$:

$$\int \frac{dx}{a^2\left(1 + \frac{x^2}{a^2}\right)} = \frac{1}{a^2}\int \frac{a\,du}{1+u^2} = \frac{1}{a}\arctan u + C = \frac{1}{a}\arctan\frac{x}{a} + C$$

This is the same idea as [trig substitution](/blog/trig-substitution), in its simplest form.

## Check by differentiating

$$\frac{d}{dx}\left(\frac{1}{a}\arctan\frac{x}{a}\right) = \frac{1}{a}\cdot\frac{1}{1 + \frac{x^2}{a^2}}\cdot\frac{1}{a} = \frac{1}{a^2 + x^2}$$

With $a = 1$ this gives back $\frac{1}{1+x^2}$.

## Definite integrals

| Integral | Value |
| --- | --- |
| $\int_0^1 \frac{dx}{1+x^2}$ | $\frac{\pi}{4} \approx 0.785$ |
| $\int_{-1}^{1} \frac{dx}{1+x^2}$ | $\frac{\pi}{2}$ |
| $\int_0^{\sqrt{3}} \frac{dx}{1+x^2}$ | $\frac{\pi}{3}$ |
| $\int_{-\infty}^{\infty} \frac{dx}{1+x^2}$ | $\pi$ |

The last one is an [improper integral](/blog/improper-integrals). It converges even though the interval is infinite, because $\frac{1}{1+x^2}$ shrinks like $\frac{1}{x^2}$.

## Worked examples

**Example 1: a constant other than 1.** $\displaystyle\int \frac{dx}{9 + x^2}$. Here $a = 3$:

$$\int \frac{dx}{9+x^2} = \frac{1}{3}\arctan\frac{x}{3} + C$$

**Example 2: a coefficient on x².** $\displaystyle\int \frac{dx}{1 + 4x^2}$. Let $u = 2x$, $du = 2\,dx$:

$$\frac{1}{2}\int \frac{du}{1+u^2} = \frac{1}{2}\arctan(2x) + C$$

**Example 3: complete the square.** $\displaystyle\int \frac{dx}{x^2 + 2x + 5}$. Write $x^2 + 2x + 5 = (x+1)^2 + 4$, then use $a = 2$ with $u = x + 1$:

$$\int \frac{dx}{(x+1)^2 + 4} = \frac{1}{2}\arctan\frac{x+1}{2} + C$$

**Example 4: split the numerator.** $\displaystyle\int \frac{x + 3}{x^2 + 1}\,dx$. Split into two integrals you know:

$$\int \frac{x}{x^2+1}\,dx + 3\int \frac{dx}{x^2+1} = \frac{1}{2}\ln(x^2+1) + 3\arctan x + C$$

The first part is covered on the [integral of x/(x²+1)](/integrals/x-over-x-squared-plus-1) page.

## Common mistakes

1. **Writing $\ln(1 + x^2)$.** That is the antiderivative of $\frac{2x}{1+x^2}$, not $\frac{1}{1+x^2}$. Without an $x$ on top, there is no $du$ for a log substitution.
2. **Dropping the $\frac{1}{a}$.** $\int \frac{dx}{4+x^2}$ is $\frac{1}{2}\arctan\frac{x}{2}$, not $\arctan\frac{x}{2}$. Differentiate your answer to catch this.
3. **Mixing it up with $\frac{1}{1-x^2}$.** A minus sign changes everything. $\frac{1}{1-x^2}$ needs [partial fractions](/blog/partial-fractions) and gives logs; see the [integral of 1/(x²−1)](/integrals/1-over-x-squared-minus-1).
4. **Using degrees.** $\arctan 1 = \frac{\pi}{4}$, not $45$. Calculus always uses radians.

For the inverse function itself, see the [arctangent](/functions/arctangent) page. To integrate $\arctan x$ (instead of getting it as an answer), see the [integral of arctan x](/integrals/arctan-x).

## Frequently asked questions

### What is the integral of 1/(1+x²)?

It is $\arctan x + C$, also written $\tan^{-1} x + C$.

### What is the integral of 1/(a²+x²)?

It is $\frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$ for $a > 0$. For example, $\int \frac{dx}{4+x^2} = \frac{1}{2}\arctan\frac{x}{2} + C$.

### What is the integral of x/(1+x²)?

It is $\frac{1}{2}\ln(1+x^2) + C$. Let $u = 1 + x^2$, so $du = 2x\,dx$. The extra $x$ on top changes the answer from an arctan to a log.

### What is the integral of 1/(1+x²) from minus infinity to infinity?

It is $\pi$. The antiderivative $\arctan x$ goes from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, so the total area is $\frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi$.