# What is int (-3x^2-5x+4 ) / (x^3-2x +1 )?

Jan 12, 2017

-ln|x^3-2x+1|-3ln|x-1|-3/2ln|x^2+x-1| -1/(2sqrt5)ln|(2x+1-sqrt5)/(2x+1+sqrt5)|+K.

#### Explanation:

Let $I = \int \frac{- 3 {x}^{2} - 5 x + 4}{{x}^{3} - 2 x + 1} \mathrm{dx} = - \int \frac{3 {x}^{2} + 5 x - 4}{{x}^{3} - 2 x + 1} \mathrm{dx}$.

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

$I = - \int \frac{3 {x}^{2} - 2}{{x}^{3} - 2 x + 1} \mathrm{dx} - \int \frac{5 x - 2}{{x}^{3} - 2 x + 1} \mathrm{dx}$

Using the Result $\int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln | f \left(x\right) | + c$, we get the First

Integral, $- \ln | {x}^{3} - 2 x + 1 | + {c}_{1.}$

Let $J = \int \frac{5 x - 2}{{x}^{3} - 2 x + 1} \mathrm{dx}$

The sum of the co-effs. of $D r .$ poly. is $0$

$\therefore D r . = \left(x - 1\right) \left({x}^{2} + x - 1\right)$.

For $J$, we will have to use the Method of Partial Fraction . So, let,

(5x-2)/{(x-1)(x^2+x-1)}=A/(x-1)+(Bx+C)/(x^2+x-1); A,B,C in RR.

The const. $A$ can be easily found by Heavyside's Method as

$A = {\left[\frac{5 x - 2}{{x}^{2} + x - 1}\right]}_{x = 1} = \frac{3}{1 + 1 - 1} = 3 ,$ &, so, by the last eqn.,

$\frac{5 x - 2}{\left(x - 1\right) \left({x}^{2} + x - 1\right)} - \frac{3}{x - 1} = \frac{B x + C}{{x}^{2} + x - 1} ,$ i.e.,

$\frac{5 x - 2 - 3 {x}^{2} - 3 x + 3}{\left(x - 1\right) \left({x}^{2} + x - 1\right)} = \frac{B x + C}{{x}^{2} + x - 1} ,$ or,

$\frac{- 3 {x}^{2} + 2 x + 1}{\left(x - 1\right) \left({x}^{2} + x - 1\right)} = \frac{\left(x - 1\right) \left(- 3 x - 1\right)}{\left(x - 1\right) \left({x}^{2} + x - 1\right)} = \frac{B x + C}{{x}^{2} + x - 1} .$

$C \le a r l y , B = - 3 , C = - 1 , \text{ &, with, } A = 3 ,$ we have,

$J = \int \frac{3}{x - 1} \mathrm{dx} - \int \frac{3 x + 1}{{x}^{2} + x - 1} \mathrm{dx}$

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

$= 3 \ln | x - 1 | + \frac{3}{2} \int \frac{\frac{d}{\mathrm{dx}} \left({x}^{2} + x - 1\right)}{{x}^{2} + x - 1} \mathrm{dx} + \frac{1}{2} \int \frac{1}{{\left(x + \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{5}}{2}\right)}^{2}} \mathrm{dx}$

$= 3 \ln | x - 1 | + \frac{3}{2} \ln | {x}^{2} + x - 1 | + \left(\frac{1}{2}\right) \left(\frac{1}{2 \frac{\sqrt{5}}{2}}\right) \ln | \frac{x + \frac{1}{2} - \frac{\sqrt{5}}{2}}{x + \frac{1}{2} + \frac{\sqrt{5}}{2}} |$

Finally, I=-ln|x^3-2x+1|-3ln|x-1|-3/2ln|x^2+x-1| -1/(2sqrt5)ln|(2x+1-sqrt5)/(2x+1+sqrt5)|+K.