Integral of 1 Over 1 Plus x Squared

In this tutorial we shall another important integration that is integral of 1 over 1 plus x squared; this integral is also using as 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 integration is the inverse process of derivative, so we have

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

As we know that 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