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

1 Answer
Oct 26, 2017

$\int \frac{- 2 {x}^{3} - x}{- 4 {x}^{2} + 2 x + 3} \mathrm{dx} = \frac{1}{4} {x}^{2} + \frac{1}{4} x + \frac{3}{8} \ln | 4 {x}^{2} - 2 x - 3 | + \frac{3 \sqrt{13}}{52} \ln | - 4 x + \sqrt{13} + 1 | - \frac{3 \sqrt{13}}{52} \ln | 4 x + \sqrt{13} - 1 | + C$

Explanation:

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

Multiply the integral by $\frac{- 1}{-} 1$:

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

Perform the implied division:

$\textcolor{w h i t e}{\frac{4 {x}^{2} - 2 x - 3}{\textcolor{b l a c k}{4 {x}^{2} - 2 x - 3}}} \frac{\frac{1}{2} x + \frac{1}{4} \textcolor{w h i t e}{x + 0}}{\text{)} \textcolor{w h i t e}{x} 2 {x}^{3} + 0 {x}^{2} + x + 0}$
$\textcolor{w h i t e}{\text{.......................}} \underline{- 2 {x}^{3} + {x}^{2} + \frac{3}{2} x}$
$\textcolor{w h i t e}{\text{.....................................}} {x}^{2} + \frac{5}{2} x$
$\textcolor{w h i t e}{\text{..................................}} \underline{- {x}^{2} + \frac{1}{2} x + \frac{3}{4}}$
$\textcolor{w h i t e}{\text{...............................................}} 3 x + \frac{3}{4}$

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

Separate into 3 integrals:

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

Integrate the first two integrals:

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

Multiply the remaining integral by $\frac{8}{8}$:

$\frac{1}{4} {x}^{2} + \frac{1}{4} x + \frac{3}{8} \int \frac{8 x + 2}{4 {x}^{2} - 2 x - 3} \mathrm{dx}$

Add to the numerator in the form of $- 2 + 2$:

$\frac{1}{4} {x}^{2} + \frac{1}{4} x + \frac{3}{8} \int \frac{8 x - 2 + 4}{4 {x}^{2} - 2 x - 3} \mathrm{dx}$

Separate into two fractions:

$\frac{1}{4} {x}^{2} + \frac{1}{4} x + \frac{3}{8} \int \frac{8 x - 2}{4 {x}^{2} - 2 x - 3} + \frac{4}{4 {x}^{2} - 2 x - 3} \mathrm{dx}$

Separate into two integrals:

$\frac{1}{4} {x}^{2} + \frac{1}{4} x + \frac{3}{8} \int \frac{8 x - 2}{4 {x}^{2} - 2 x - 3} \mathrm{dx} + \frac{3}{8} \int \frac{4}{4 {x}^{2} - 2 x - 3} \mathrm{dx}$

The first integral is the natural logarithm:

$\frac{1}{4} {x}^{2} + \frac{1}{4} x + \frac{3}{8} \ln | 4 {x}^{2} - 2 x - 3 | + \frac{3}{2} \int \frac{1}{4 {x}^{2} - 2 x - 3} \mathrm{dx}$

The last integral is done by completing the square:

$\frac{1}{4} {x}^{2} + \frac{1}{4} x + \frac{3}{8} \ln | 4 {x}^{2} - 2 x - 3 | + \frac{3 \sqrt{13}}{52} \ln | - 4 x + \sqrt{13} + 1 | - \frac{3 \sqrt{13}}{52} \ln | 4 x + \sqrt{13} - 1 | + C$