# What is int (x^3-2x^2+5x-3 ) / (-x^2+ 3 x -4 )dx?

Dec 11, 2017

$- \left({x}^{2} / 2 + x\right) - 2 \ln | {x}^{2} - 3 x + 4 | + \frac{2}{\sqrt{7}} \cdot a r c \tan \left(\frac{2 x - 3}{\sqrt{7}}\right) + C .$

#### Explanation:

Let, $I = \int \frac{{x}^{3} - 2 {x}^{2} + 5 x - 3}{- {x}^{2} + 3 x - 4} \mathrm{dx} ,$

$= - \int \frac{{x}^{3} - 2 {x}^{2} + 5 x - 3}{{x}^{2} - 3 x + 4} \mathrm{dx} .$

Now, ${x}^{3} - 2 {x}^{2} + 5 x - 3 ,$

$= x \left({x}^{2} - 3 x + 4\right) + 1 \left({x}^{2} - 3 x + 4\right) + 4 x - 7 ,$

$= \left(x + 1\right) \left({x}^{2} - 3 x + 4\right) + \left(4 x - 7\right) .$

$\Rightarrow \frac{{x}^{3} - 2 {x}^{2} + 5 x - 3}{{x}^{2} - 3 x + 4} ,$

$= x + 1 + \frac{4 x - 7}{{x}^{2} - 3 x + 4} .$

$\therefore I = - \int \left\{x + 1 + \frac{4 x - 7}{{x}^{2} - 3 x + 4}\right\} \mathrm{dx} ,$

$= - \left({x}^{2} / 2 + x\right) - \int \frac{4 x - 7}{{x}^{2} - 3 x + 4} \mathrm{dx} \ldots \ldots . \left(\star\right) .$

To evaluate the Integral in the R.H.S. of $\left(\star\right) ,$ we have to set,

4x-7=m*d/dx(x^2-3x+4)+n," where "m,n in RR; i.e.,

we have to determine $m , n \in \mathbb{R} ,$ such that,

$4 x - 7 = m \left(2 x - 3\right) + n .$

Clearly, $m = 2 , n = - 1.$

$\therefore \int \frac{4 x - 7}{{x}^{2} - 3 x + 4} \mathrm{dx} ,$

$= \int \frac{2 \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} - 3 x + 4\right) - 1}{{x}^{2} - 3 x + 4} \mathrm{dx} ,$

$= 2 \int \frac{\frac{d}{\mathrm{dx}} \left({x}^{2} - 3 x + 4\right)}{{x}^{2} - 3 x + 4} \mathrm{dx} - \int \frac{1}{{x}^{2} - 3 x + 4} \mathrm{dx} ,$

$= 2 \ln | \left({x}^{2} - 3 x + 4\right) | - \int \frac{1}{{x}^{2} - 3 x + \frac{9}{4} + \frac{7}{4}} \mathrm{dx} ,$

$= 2 \ln | \left({x}^{2} - 3 x + 4\right) | - \int \frac{1}{{\left(x - \frac{3}{2}\right)}^{2} + {\left(\frac{\sqrt{7}}{2}\right)}^{2}} \mathrm{dx} ,$

$= 2 \ln | {x}^{2} - 3 x + 4 | - \frac{1}{\frac{\sqrt{7}}{2}} \cdot a r c \tan \left\{\frac{x - \frac{3}{2}}{\frac{\sqrt{7}}{2}}\right\} .$

Altogether, we have,

$I = - \left({x}^{2} / 2 + x\right) - 2 \ln | {x}^{2} - 3 x + 4 |$
$+ \frac{2}{\sqrt{7}} \cdot a r c \tan \left(\frac{2 x - 3}{\sqrt{7}}\right) + C .$

Enjoy Maths.!