Integral of 1 Over 1 Plus x Squared

In this tutorial we shall look at another important integration, the integral of $1$ over $1$ plus $x$ squared. This integral also uses an important formula.

The integral of $\frac{1}{{1 + {x^2}}}$ is of the form

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

To prove this formula, consider

$$\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}x + c} \right] = \frac{d}{{dx}}{\tan ^{ - 1}}x + \frac{d}{{dx}}c$$

Using the derivative formula $\frac{d}{{dx}}{\tan ^{ - 1}}x = \frac{1}{{1 + {x^2}}}$, we have

$$\begin{gathered} \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}x + c} \right] = \frac{1}{{1 + {x^2}}} + 0 \\ \Rightarrow \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}x + c} \right] = \frac{1}{{1 + {x^2}}} \\ \Rightarrow \frac{1}{{1 + {x^2}}} = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}x + c} \right] \\ \Rightarrow \frac{1}{{1 + {x^2}}}dx = d\left[ {{{\tan }^{ - 1}}x + c} \right]\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right) \\ \end{gathered}$$

Integrating both sides of equation (i) with respect to $x$, we have

$$\int {\frac{1}{{1 + {x^2}}}dx} = \int {d\left[ {{{\tan }^{ - 1}}x + c} \right]}$$

As we know that by definition the integration is the inverse process of the derivative, we have

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

We know that the derivative of

$$\frac{d}{{dx}}{\cot ^{ - 1}}x = - \frac{1}{{1 + {x^2}}}$$

This formula can also be written as

$$\int {\frac{1}{{1 + {x^2}}}dx} = - {\cot ^{ - 1}}x + c$$