What is int (2x^3-3x^2-4x-3 ) / (-6x^2+ 3 x -4 )dx?

1 Answer
Mar 10, 2016

$- {x}^{2} / 6 + \frac{x}{3} + \frac{19}{18} \ln | \frac{\frac{1}{2} \sqrt{\frac{29}{2}}}{\sqrt{6 {x}^{2} - 3 x + 4}} | - \frac{13}{2 \sqrt{87}} \cdot {\tan}^{- 1} \left(4 \sqrt{\frac{3}{29}} \left(x - \frac{1}{4}\right)\right) + c o n s t .$

Explanation:

First we should get in the numerator a polynomial of a grade inferior than the denominator's

By long division:

$\text{ "2x^3-3x^2-4x-3" }$|$\text{ } - 6 {x}^{2} + 3 x - 4$
$- 2 {x}^{3} + {x}^{2} - \frac{4}{3} x \text{ }$|____
_____" "-1/3x+1/3
$\text{ } - 2 {x}^{2} - \frac{16}{3} x - 3$
$\text{ } 2 {x}^{2} - x + \frac{4}{3}$
$\text{ }$ _______
$\text{ } - \frac{19}{3} x - \frac{5}{3} = - \frac{1}{3} \left(19 x + 5\right)$

So the expression becomes
$= \frac{1}{3} \int \left(- x + 1\right) \mathrm{dx} - \frac{1}{3} \cdot \frac{1}{6} \int \frac{19 x + 5}{{x}^{2} - \frac{x}{2} + \frac{2}{3}} \mathrm{dx}$

Dealing with
$= \int \frac{19 x + 5}{{x}^{2} - \frac{x}{2} + \frac{2}{3}} \mathrm{dx}$
Since ${\left(x - \frac{1}{4}\right)}^{2} = {x}^{2} - \frac{x}{2} + \frac{1}{16}$ and $\frac{2}{3} - \frac{1}{16} = \frac{29}{48} = {\left(\frac{1}{4} \sqrt{\frac{29}{3}}\right)}^{2}$
We can make

$x - \frac{1}{4} = \frac{1}{4} \sqrt{\frac{29}{3}} \cdot \tan y$
=> $\mathrm{dx} = \frac{1}{4} \sqrt{\frac{29}{3}} \cdot {\sec}^{2} y \cdot \mathrm{dy}$
Now let's find how many units of $\left(x - \frac{1}{4}\right)$ there are in the denominator
$\text{ "19x+5" }$|$\text{ } x - \frac{1}{4}$
$- 19 x + \frac{19}{4} \text{ }$|____
_____" "19
$\text{ } \frac{39}{4}$

So the partial expression becomes
$= 19 \int \frac{\left(\frac{1}{4} \sqrt{\frac{29}{3}} \tan y\right) \left(\frac{1}{4} {\sec}^{2} y\right)}{\frac{1}{16} \cdot \frac{29}{3} {\sec}^{2} y} \mathrm{dy} + \frac{39}{4} \int \frac{\frac{1}{4} \sqrt{\frac{29}{3}}}{\frac{1}{16} \cdot \frac{29}{3} {\sec}^{2} y} \mathrm{dy}$
$= - 19 \ln | \cos y | + 39 \sqrt{\frac{3}{29}} y$
But $\tan y = \sqrt{\frac{3}{29}} \left(4 x - 1\right)$ => $\sin y = \sqrt{\frac{3}{29}} \left(4 x - 1\right) \cos y$
And ${\sin}^{2} y + {\cos}^{2} y = 1$ => $\left(\frac{3}{29} \left(16 {x}^{2} - 8 x + 1\right) + 1\right) {\cos}^{2} y = 1$ => $\frac{48 {x}^{2} - 24 x + 32}{29} \cdot {\cos}^{2} y = 1$ => cos y=(1/2sqrt(29/2))/sqrt(6x^2-4x+8
Then the partial expression becomes
$= - 19 \ln | \frac{\frac{1}{2} \sqrt{\frac{29}{2}}}{\sqrt{6 {x}^{2} - 4 x + 8}} + 39 \sqrt{\frac{3}{29}} {\tan}^{- 1} \left(\sqrt{\frac{3}{29}} \left(4 x - 1\right)\right)$

Back to the main expression:
$= - {x}^{2} / 6 + \frac{1}{3} - \frac{1}{18} \left[- 19 \ln | \frac{\frac{1}{2} \sqrt{\frac{29}{2}}}{\sqrt{6 {x}^{2} - 4 x + 8}} + 39 \sqrt{\frac{3}{29}} {\tan}^{- 1} \left(\sqrt{\frac{3}{29}} \left(4 x - 1\right)\right)\right]$
$= - {x}^{2} / 6 + \frac{x}{3} + \frac{19}{18} \ln | \frac{\frac{1}{2} \sqrt{\frac{29}{2}}}{\sqrt{6 {x}^{2} - 3 x + 4}} | - \frac{13}{2 \sqrt{87}} \cdot {\tan}^{- 1} \left(4 \sqrt{\frac{3}{29}} \left(x - \frac{1}{4}\right)\right) + c o n s t .$

One step further: simplifying
The result above is good enough, but the constant part of the logarithm can be absorbed by the general constant:
$= - {x}^{2} / 6 + \frac{x}{3} + \frac{19}{18} \left[\ln \left(\frac{1}{2} \sqrt{\frac{29}{2}}\right) - \left(\frac{1}{2}\right) \ln | 6 {x}^{2} - 3 x + 4 |\right] - \frac{13}{2 \sqrt{87}} \cdot {\tan}^{- 1} \left(4 \sqrt{\frac{3}{29}} \left(x - \frac{1}{4}\right)\right) + c o n s t .$

$= - {x}^{2} / 6 + \frac{x}{3} - \frac{19}{36} \ln | 6 {x}^{2} - 3 x + 4 | - \frac{13}{2 \sqrt{87}} \cdot {\tan}^{- 1} \left(4 \sqrt{\frac{3}{29}} \left(x - \frac{1}{4}\right)\right) + c o n s t .$