Problems on Techniques of Integration

We have

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

So

\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}

Detailed Answer.


If you prefer to jump to the next problem, click on Next Problem below.

[Next Problem] [Matrix Algebra]
[Trigonometry] [Calculus]
[Geometry] [Algebra]
[Differential Equations] [Complex Variables]

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Mohamed A. Khamsi

Copyright 1999-2017 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour