Question

Find the indefinite integral \[\int\limits_{}\tan^{-1}xdx = \int\limits_{}arctanxdx\]

Answer 1

There we go, I got something ^.^

Answer 2

writing \[u=\tan^{-1}x \] and \[v = x\] and using the integration by parts we have find, \[du = (1) / ( x^2 + 1 )\] and v = r

Answer 3

so lets roll ^.^

Answer 4

\[\int\limits_{}\tan^{-1}xdx =(\tan^{-1}x) - \int\limits_{}(r)( 1 / (1 + r^2))dr\]
\[=x \tan^{-1}x - (1/2)\int\limits_{}(2xdx)/(1 + r^2)\]
\[=x \tan^{-1}x - (1/2)\int\limits_{}dy/y\] setting \[y = r^2 + 1\]
...
= \[x \tan^{-1}x - (1/2)\ln|y| + c \]
=\[x \tan^{-1}x - (1/2)\ln(r^2 + 1) + c \]
since \[y = r^2 + 1 > 0\]