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.

Set

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

Then

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

Since

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

we get

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

In order to integrate the function $\displaystyle \frac{x^2}{x^2 + 1}$, we will need the technique of integrating rational functions. But in this case, we may use the identity

\begin{displaymath}\frac{x^2}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}\end{displaymath}

which implies

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

So

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

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

Short Answer.

Detailed Answer.


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