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