# What is int (-2x^3-x ) / (-4x^2+2x +7 )?

Feb 17, 2018

$\int \frac{- 2 {x}^{3} - x}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx}$
$= 0.75 x + 0.1875 \ln \left(4 {x}^{2} - 2 x - 7\right) - 1.637 \ln \left(\frac{x + 1.096}{x - 1.596}\right)$

#### Explanation:

To find:
$\int \frac{- 2 {x}^{3} - x}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx}$

Let $I = \int \frac{- 2 {x}^{3} - x}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx}$

Dividing the fraction in the integrand

We get
$\left(- 4 {x}^{2} + 2 x + 7\right) \times 0.5 x = - 2 {x}^{3} + {x}^{2} + 3.5$

$- 2 {x}^{3} - x = - 2 {x}^{3} + {x}^{2} + 3.5 - {x}^{2} - 3.5 - x$
Thus,

$I = \int \frac{- 2 {x}^{3} + {x}^{2} + 3.5 - {x}^{2} - 3.5 - x}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx}$

$\int \frac{- 2 {x}^{3} + {x}^{2} + 3.5}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx} + \int \frac{- {x}^{2} - 3.5 - x}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx}$

Ler ${I}_{1} = \int \frac{- 2 {x}^{3} + {x}^{2} + 3.5}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx} \mathmr{and} {I}_{2} = \int \frac{- {x}^{2} - 3.5 - x}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx}$

Then,
$I = {I}_{1} + {I}_{2}$

${I}_{1} = \int \frac{- 2 {x}^{3} + {x}^{2} + 3.5}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx} = \int 0.5 \mathrm{dx} = 0.5 x$

${I}_{1} = 0.5 x$

${I}_{2} = \int \frac{- {x}^{2} - 3.5 - x}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx}$

Taking negative sign, and rearranging the numerator

${I}_{2} = \int \frac{{x}^{2} + x + 3.5}{4 {x}^{2} - 2 x - 7} \mathrm{dx}$

Now, the numerator can be expressed as a combination

$\left({x}^{2} + x + 3.5\right) = p \left(4 {x}^{2} - 2 x - 7\right) + q \frac{d}{\mathrm{dx}} \left(4 {x}^{2} - 2 x - 7\right) + r$

$\frac{d}{\mathrm{dx}} \left(4 {x}^{2} - 2 x - 7\right) = 8 x - 2$

Thus,
${x}^{2} + x + 3.5 = p \left(4 {x}^{2} - 2 x - 7\right) + q \left(8 x - 2\right) + r$

Simplifying

${x}^{2} + x + 3.5 = 4 p {x}^{2} - 2 p x - 7 p + 8 q x - 2 q + r$

Collecting the terms having like powers of x

${x}^{2} + x + 3.5 = 4 p {x}^{2} + \left(- 2 p + 8 q\right) x + \left(- 7 p - 2 q + r\right)$

Equating the coefficients of like powers of x

$1 = 4 p$
$p = \frac{1}{4} = 0.25$

$1 = - 2 p + 8 q$
$1 = - 2 \times \frac{1}{4} + 8 q$
$1 = - \frac{1}{2} + 8 q$
$\frac{3}{2} = 8 q$
$q = \frac{3}{16} = 0.1875$

$3.5 = - 7 p - 2 q + r$
$3.5 = - 7 \times 0.25 - 2 \times 0.1875 + r$
$3.5 = - 1.75 - 0.375 + r$
$3.5 = - 2.125 + r$

$r = 3.5 + 2.125$

$r = 5.625$

Thus, we have,
$p = 0.25 , q = 0.1875 , \mathmr{and} r = 5.625$

$p \left(4 {x}^{2} - 2 x - 7\right) + q \left(8 x - 2\right) + r = 0.25 \left(4 {x}^{2} - 2 x - 7\right) + 0.1875 \left(8 x - 2\right) + 5.625$

and
$\int \frac{{x}^{2} + x + 3.5}{4 {x}^{2} - 2 x - 7} \mathrm{dx} = \int \left(0.25 \left(4 {x}^{2} - 2 x - 7\right) + 0.1875 \left(8 x - 2\right) + \frac{5.625}{4 {x}^{2} - 2 x - 7}\right) \mathrm{dx}$

Using the sum rule and simplifying

${I}_{2} = 0.25 \int \frac{4 {x}^{2} - 2 x - 7}{4 {x}^{2} - 2 x - 7} \mathrm{dx} + 0.1875 \int \frac{\left(8 x - 2\right) \mathrm{dx}}{4 {x}^{2} - 2 x - 7} + 5.625 \int \frac{1}{4 {x}^{2} - 2 x - 7} \mathrm{dx}$

$0.25 \int \frac{4 {x}^{2} - 2 x - 7}{4 {x}^{2} - 2 x - 7} \mathrm{dx} = 0.25 \int 1 \mathrm{dx} = 0.25 x$
$0.1875 \int \frac{2 x - 2}{4 {x}^{2} - 2 x - 7} \mathrm{dx} =$

Let $t = 4 {x}^{2} - 2 x - 7 , \mathrm{dt} = \left(8 x - 2\right) \mathrm{dx}$

Then
$0.1875 \int \frac{2 x - 2}{4 {x}^{2} - 2 x - 7} \mathrm{dx} = 0.1875 \int \frac{\mathrm{dt}}{t} = 0.1875 \ln t$
Substituting for t

$0.1875 \int \frac{2 x - 2}{4 {x}^{2} - 2 x - 7} \mathrm{dx} = 0.1875 \ln \left(4 {x}^{2} - 2 x - 7\right)$

$5.625 \int \frac{1}{4 {x}^{2} - 2 x - 7} \mathrm{dx} =$

Completing the squares in the denominator

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

$= 4 \left({\left(x - \frac{1}{4}\right)}^{2} - \left(\frac{7}{4} + \frac{1}{16}\right)\right)$
$= 4 \left({\left(x - \frac{1}{4}\right)}^{2} - 1.8125\right)$
$1.8125 = {1.346}^{2}$

Thus,
$5.625 \int \frac{1}{4 {x}^{2} - 2 x - 7} \mathrm{dx} = 5.625 \int \frac{1}{4 \left({\left(x - \frac{1}{4}\right)}^{2} - {1.346}^{2}\right)} \mathrm{dx}$

$L e t t = x - \frac{1}{4} , \mathrm{dt} = \mathrm{dx}$

Now,

$5.625 \int \frac{1}{4 {x}^{2} - 2 x - 7} \mathrm{dx} = \frac{5.625}{4} \int \frac{\mathrm{dt}}{{t}^{2} - {1.346}^{2}} \mathrm{dx}$

$\frac{5.625}{4} = 1.40625$
$\int \frac{\mathrm{dt}}{{t}^{2} - {1.346}^{2}} = \int \frac{\mathrm{dt}}{\left(t + 1.346\right) \left(t - 1.346\right)}$

$\int \frac{\mathrm{dx}}{\left(x + a\right) \left(x + b\right)} = \frac{1}{b - a} \ln \left(\frac{a + x}{b + x}\right) , a \ne b$

We have,

$x = t , a = 1.346 , b = - 1.346 , b - a = - 1.346 - 1.346 = - 2.692$

Thus,
$\int \frac{\mathrm{dt}}{\left(t + 1.346\right) \left(t - 1.346\right)} = \frac{1}{- 2.692} \ln \left(\frac{1.346 + t}{- 1.346 + t}\right)$

Substituting for t
$1.346 + t = 1.346 + \left(x - \frac{1}{4}\right) = 1.346 + x - 0.25 = x + 1.096$

$- 1.346 + t = - 1.346 + \left(x - \frac{1}{4}\right) = - 1.346 + x - 0.25 = x - 1.596$

$\int \frac{\mathrm{dt}}{\left(t + 1.346\right) \left(t - 1.346\right)} = \frac{1}{- 2.692} \ln \left(\frac{x + 1.096}{x - 1.596}\right)$

Now,

$5.625 \int \frac{1}{4 {x}^{2} - 2 x - 7} \mathrm{dx}$
$= 1.40625 \times \frac{1}{- 2.692} \ln \left(\frac{x + 1.096}{x - 1.596}\right)$
$= - 1.637 \ln \left(\frac{x + 1.096}{x - 1.596}\right)$

Thus,

${I}_{1} = 0.5 x$

${I}_{2} = \int \frac{- {x}^{2} - 3.5 - x}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx}$

$= 0.25 x$
$+ 0.1875 \ln \left(4 {x}^{2} - 2 x - 7\right)$
$+ \left(- 1.637 \ln \left(\frac{x + 1.096}{x - 1.596}\right)\right)$

${I}_{2} = 0.25 x + 0.1875 \ln \left(4 {x}^{2} - 2 x - 7\right) - 1.637 \ln \left(\frac{x + 1.096}{x - 1.596}\right)$

I=I_1+I_2

$\int \frac{- 2 {x}^{3} - x}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx}$
$= 0.5 x + 0.25 x + 0.1875 \ln \left(4 {x}^{2} - 2 x - 7\right) - 1.637 \ln \left(\frac{x + 1.096}{x - 1.596}\right)$

$\int \frac{- 2 {x}^{3} - x}{- 4 {x}^{2} + 2 x + 7} \mathrm{dx}$
$= 0.75 x + 0.1875 \ln \left(4 {x}^{2} - 2 x - 7\right) - 1.637 \ln \left(\frac{x + 1.096}{x - 1.596}\right)$