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 $\arctan(x)$.

Set

\begin{displaymath}\left\{\begin{array}{lll}
u &=& \arctan(2 x)\\
dv &=& dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=&\displaystyle \frac{2 }{4 x^2 + 1} dx\\
v &=& x\;.
\end{array}\right.\end{displaymath}

Since

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

we get So

\begin{displaymath}\int \arctan(2 x) dx = x \arctan(2 x) - \int x \frac{2}{4 x^2 + 1} dx = x \arctan(x) - \int \frac{2 x}{4 x^2 + 1}dx\end{displaymath}

But

\begin{displaymath}\int \frac{2 x}{4 x^2 + 1}dx = \int \frac{1}{4}\; \frac{8 x}{4 x^2 + 1}dx = \frac{1}{4} \ln(4 x^2+1)\end{displaymath}

which implies

\begin{displaymath}\int \arctan(2 x) dx = x \arctan(2 x) - \frac{1}{4}\ln(4 x^2+1) + C\;.\end{displaymath}

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


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