Blog: `int (tan^(-1) (x))/(1 + x^2) dx` Evaluate the indefinite integral.

Monday, October 29, 2012

You need to use the following substitution `tan^(-1) x = t` , such that:

`tan^(-1) x= t => (dx)/(1+x^2) = dt`

`int ((tan^(-1) x)dx)/(1+x^2) = int t dt`

`int t dt = t^2/2 + c`

Replacing back `tan^(-1) x` for t yields:

`int ((tan^(-1) x)dx)/(1+x^2) = ((tan^(-1) x)^2)/2 + c`

Hence, evaluating the indefinite integral, yields `int ((tan^(-1) x)dx)/(1+x^2) = ((tan^(-1) x)^2)/2 + c.`

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