Integration of Tangent Inverse

The tangent inverse function {\tan ^{ - 1}}x it is an important integral function, but it has no direct method to find it. We shall find the integration of tangent inverse by using the integration by parts method.

The integration of tangent inverse is of the form

I = \int {{{\tan }^{ - 1}}xdx}

To solve this integration, it must have at least two functions, however it has only one function: {\tan ^{ - 1}}x. So, consider the second function as 1. Now the integration becomes

I = \int {{{\tan }^{ - 1}}x \cdot 1dx} \,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

The first function is {\tan ^{ - 1}}x and the second function is 1.

Using the formula for integration by parts, we have

\int {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } }

Using the formula above equation (i) becomes

\begin{gathered} I = {\tan ^{ - 1}}x\int {1dx - \int {\left[ {\frac{d}{{dx}}{{\tan }^{ - 1}}x\int {1dx} } \right]dx} } \\ \Rightarrow I = x{\tan ^{ - 1}}x - \int {\left[ {\frac{1}{{1 + {x^2}}}x} \right]dx} \\ \Rightarrow I = x{\tan ^{ - 1}}x - \int {\frac{x}{{1 + {x^2}}}dx} \\ \end{gathered}

Multiplying and dividing by 2, we have

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

Using formula

\int {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = \ln f\left( x \right)} + c

we have

\begin{gathered} I = x{\tan ^{ - 1}}x - \frac{1}{2}\ln \left( {1 + {x^2}} \right) + c \\ \Rightarrow \int {{{\tan }^{ - 1}}xdx} = x{\tan ^{ - 1}}x - \frac{1}{2}\ln \left( {1 + {x^2}} \right) + c \\ \end{gathered}

Now we can also use this integration of tangent inverse as a formula.