# Evaluate the integral  int \ (4x^3-7x)/(x^4-5x^2+4) \ dx ?

Jul 29, 2017

$\int \setminus \frac{4 {x}^{3} - 7 x}{{x}^{4} - 5 {x}^{2} + 4} \setminus \mathrm{dx} = \frac{3}{2} \ln | {x}^{2} - 4 | + \frac{1}{2} \ln | {x}^{2} - 1 | + c$

#### Explanation:

Denote the integral by:

$I = \int \setminus \frac{4 {x}^{3} - 7 x}{{x}^{4} - 5 {x}^{2} + 4} \setminus \mathrm{dx}$

The denominator of the integrand is quadratic in ${x}^{2}$ and factorises as follows:

$\frac{4 {x}^{3} - 7 x}{{x}^{4} - 5 {x}^{2} + 4} \equiv \frac{4 {x}^{3} - 7 x}{\left({x}^{2} - 4\right) \left({x}^{2} - 1\right)}$
$\text{ } = \frac{4 {x}^{3} - 7 x}{\left({x}^{2} - {2}^{2}\right) \left({x}^{2} - {1}^{2}\right)}$
$\text{ } = \frac{4 {x}^{3} - 7 x}{\left(x - 2\right) \left(x + 2\right) \left(x - 1\right) \left(x + 1\right)}$

And so the partial fraction decomposition will be of the form:

$\frac{4 {x}^{3} - 7 x}{{x}^{4} - 5 {x}^{2} + 4} \equiv \frac{A}{x - 2} + \frac{B}{x + 2} + \frac{C}{x - 1} + \frac{D}{x - 1}$

Then if we put the RHS over a common denominator we end up with the identity:

$\left(4 {x}^{3} - 7 x\right) \equiv A \left(x + 2\right) \left(x - 1\right) \left(x + 1\right) + B \left(x - 2\right) \left(x - 1\right) \left(x + 1\right) + C \left(x - 2\right) \left(x + 2\right) \left(x + 1\right) + D \left(x - 2\right) \left(x + 2\right) \left(x - 1\right)$

Where $A , B , C , D$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put $x = \setminus \setminus \setminus \setminus 2 \implies 4 \cdot 8 - 14 = A \left(4\right) \left(1\right) \left(3\right) \implies A = \frac{3}{2}$
Put $x = - 2 \implies - 4 \cdot 8 + 14 = B \left(- 4\right) \left(- 3\right) \left(- 1\right) \implies B = \frac{3}{2}$
Put $x = \setminus \setminus \setminus \setminus 1 \implies 4 - 7 = C \left(- 1\right) \left(3\right) \left(2\right) \implies C = \frac{1}{2}$
Put $x = - 1 \implies - 4 + 7 = D \left(- 3\right) \left(1\right) \left(- 2\right) \implies D = \frac{1}{2}$

So using partial fraction decomposition we have:

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

These are all trivial integrals, so we can integrate to get:

$I = \frac{3}{2} \ln | x - 2 | + \frac{3}{2} \ln | x + 2 | + \frac{1}{2} \ln | x - 1 | + \frac{1}{2} \ln | x - 1 | + c$
$\setminus \setminus = \frac{3}{2} \left(\ln | x - 2 | + \ln | x + 2 |\right) + \frac{1}{2} \left(\ln | x - 1 | + \ln | x - 1 |\right) + c$
$\setminus \setminus = \frac{3}{2} \left(\ln | x - 2 | | x + 2 |\right) + \frac{1}{2} \left(\ln | x - 1 | | x - 1 |\right) + c$
$\setminus \setminus = \frac{3}{2} \left(\ln | \left(x - 2\right) \left(x + 2\right) |\right) + \frac{1}{2} \left(\ln | \left(x - 1\right) \left(x - 1\right) |\right) + c$
$\setminus \setminus = \frac{3}{2} \ln | {x}^{2} - 4 | + \frac{1}{2} \ln | {x}^{2} - 1 | + c \setminus \setminus \setminus \setminus \setminus$ QED