Integration of Tangent Inverse

Tangent inverse function {\tan ^{ - 1}}x it is an important integral function, but it has no direct method to find we shall find the integration of tangent inverse by using 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, it in this function it has only one function that is {\tan ^{ - 1}}x, now consider second function as 1. Now the integration becomes

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


Take first function is {\tan  ^{ - 1}}x and second function will be 1
Using 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} } }


Equation (i) becomes using above formula, we have

\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 further we can use this integration of tangent inverse as a formula.

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