Problems on Techniques of Integration

The Arctangent function is one of these functions whose derivative is very nice (rational function). So it is a good idea to use the integration by parts technique in which $\arctan(x)$ will be differentiated. One may argue that we do not have two functions to apply this technique. That's true except that $\arctan(x)$ may also be seen as the product of the functions $1$ and $\arctan(x)$.


u &=& \arctan(x)\\
dv &=& dx\;.


du &=&\displaystyle \frac{1}{x^2 + 1} dx\\
v &=& x\;.


\begin{displaymath}\int u dv = u v - \int v du\;,\end{displaymath}

we get So

\begin{displaymath}\int \arctan(x) dx = x \arctan(x) - \int x \frac{1}{x^2 + 1} dx = x \arctan(x) - \int \frac{x}{x^2 + 1}dx\end{displaymath}


\begin{displaymath}\int \frac{x}{x^2 + 1}dx = \int \frac{1}{2}\; \frac{2x}{x^2 + 1}dx = \frac{1}{2} \ln(x^2+1)\end{displaymath}

which implies

\begin{displaymath}\int \arctan(x) dx = x \arctan(x) - \frac{1}{2}\ln(x^2+1) + C\;.\end{displaymath}

It is a common mistake to forget the constant $C$.

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Mohamed A. Khamsi

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