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
\[\int {\frac{1}{{1 + {x^2}}}dx} = – {\cot ^{ – 1}}x + c\]