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

May 6, 2018

$I = \frac{3 {x}^{2}}{4} - \frac{3 x}{2} + \frac{9}{8} \ln | {x}^{2} + x - \frac{1}{2} | - \frac{31}{8 \sqrt{3}} \ln | \frac{2 x + 1 - \sqrt{3}}{2 x + 1 + \sqrt{3}} | + C$

#### Explanation:

Here,

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

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

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

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

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

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

$= \frac{3}{2} \cdot {x}^{2} / 2 - \frac{3}{2} \int \frac{\left({x}^{2} + x - \frac{1}{2}\right) - \left(\frac{3 x}{2} - \frac{11}{6}\right)}{{x}^{2} + x - \frac{1}{2}} \mathrm{dx}$

$= \frac{3 {x}^{2}}{4} - \frac{3}{2} \int 1 \mathrm{dx} + \frac{3}{2} \int \frac{\frac{3 x}{2} - \frac{11}{6}}{{x}^{2} + x - \frac{1}{2}} \mathrm{dx}$

$= \frac{3 {x}^{2}}{4} - \frac{3 x}{2} + \frac{3}{2} \times \frac{3}{2} \int \frac{x - \frac{11}{9}}{{x}^{2} + x - \frac{1}{2}} \mathrm{dx}$

$= \frac{3 {x}^{2}}{4} - \frac{3 x}{2} + \frac{9}{4} \times \frac{1}{2} \int \frac{2 x - \frac{22}{9}}{{x}^{2} + x - \frac{1}{2}} \mathrm{dx}$

$= \frac{3 {x}^{2}}{4} - \frac{3 x}{2} + \frac{9}{8} \int \frac{\left(2 x + 1\right) - \frac{31}{9}}{{x}^{2} + x - \frac{1}{2}} \mathrm{dx}$

=(3x^2)/4-(3x)/2+9/8int(2x+1)/(x^2+x-1/2)dx- 9/8int(31/9)/(x^2+x-1/2)dx/

$= \frac{3 {x}^{2}}{4} - \frac{3 x}{2} + \frac{9}{8} \ln | {x}^{2} + x - \frac{1}{2} | - \frac{9}{8} \times \frac{31}{9} {I}_{1.} . . \to \left(A\right)$

Where, I_1=int1/(x^2+x+1/4-3/4)dx=int1/((x^+1/2)^2- (sqrt3/2))dx

$\therefore {I}_{1} = \frac{1}{2 \left(\frac{\sqrt{3}}{2}\right)} \ln | \frac{x + \frac{1}{2} - \frac{\sqrt{3}}{2}}{x + \frac{1}{2} + \frac{\sqrt{3}}{2}} | + c$

${I}_{1} = \frac{1}{\sqrt{3}} \ln | \frac{2 x + 1 - \sqrt{3}}{2 x + 1 + \sqrt{3}} | + c$

Hence, from $\left(A\right)$,

I=(3x^2)/4-(3x)/2+9/8ln|x^2+x-1/2|-31/(8sqrt3)ln|(2x+1- sqrt3)/(2x+1+sqrt3)|+C