Problems on Techniques of Integration

The answer is given in any Table of integrals. Here we try to give the details for getting the answer.

By completing the square, we have

\begin{displaymath}\begin{array}{lll}
ax^2 + b x + c &=&\displaystyle a \left(x^...
...ac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a^2} \right)
\end{array}\end{displaymath}

Since $b^2 - 4a c < 0$, set

\begin{displaymath}\omega = \frac{\sqrt{4ac - b^2}}{2a} \cdot\end{displaymath}

So

\begin{displaymath}\int \frac{1}{ax^2 + b x + c}dx = \frac{1}{a} \int \frac{1}{\displaystyle \left(x + \frac{b}{2a}\right)^2 + \omega^2}dx\cdot\end{displaymath}

Hence

\begin{displaymath}\int \frac{1}{ax^2 + b x + c}dx = \frac{1}{a \omega} \arctan\left(\frac{x + b/2a}{\omega}\right) + C \end{displaymath}

or

\begin{displaymath}\int \frac{1}{ax^2 + b x + c}dx = \frac{2}{\sqrt{4ac - b^2}} \arctan\left(\frac{2a x + b}{\sqrt{4ac - b^2}}\right) + C .\end{displaymath}

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

Remark. Recall the $\arctan$ formula

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


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Mohamed A. Khamsi

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