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. We have

\begin{displaymath}x = \frac{1}{2a} \Big(2a x + b) - \frac{b}{2a}\cdot\end{displaymath}


\begin{displaymath}\int \frac{x}{ax^2 + b x + c}dx = \frac{1}{2a} \int \frac{2ax...
... + b x + c}dx - \frac{b}{2a} \int \frac{1}{ax^2 + b x + c}\cdot\end{displaymath}

On one hand, we have

\begin{displaymath}\int \frac{2ax+b}{ax^2 + b x + c}dx = \ln\vert ax^2 + b x + c\vert,\end{displaymath}

and also from Table or previous problem, we get

\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).\end{displaymath}

We get

\begin{displaymath}\int \frac{x}{ax^2 + b x + c}dx = \frac{1}{2a} \ln\vert ax^2 ...
...^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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