# Use the method of partial fraction decomposition to perform the required integration. \int (x^2 -...

## Question:

Use the method of partial fraction decomposition to perform the required integration.

{eq}\int \frac{x^{3} - 3x}{(x^{2} + 6)^{2}} dx {/eq}

## Partial Fraction Decomposition

The partial fraction decomposition, also called the partial fraction expansion, of a rational function is a way of writing the rational function as a sum of simpler rational functions. The writing of a rational function as its partial fraction decomposition is essentially the reverse procedure of combining multiple rational expressions together by finding a common denominator and adding. To find the partial fraction decomposition, begin by fully factoring the denominator. For each linear factor {eq}ax+b {/eq} in the denominator, there is a corresponding term {eq}\dfrac{A}{ax+b} {/eq} in the partial fraction decomposition. Similarly, for every irreducible quadratic factor in the denominator {eq}ax^2+bx+c {/eq} there is a corresponding term in the partial fraction decomposition of the form {eq}\dfrac{Ax+B}{ax^2+bx+c} {/eq}. Once a rational function is fully decomposed into partial fractions, it may be easier to do other things with the function, such as integration.

To evaluate {eq}\displaystyle\int \dfrac{x^{3} - 3x}{(x^{2} + 6)^{2}} dx {/eq}, begin by finding the partial fraction decomposition of the integrand. The denominator is already factored as a power of an irreducible quadratic factor. So we have

{eq}\dfrac{x^{3} - 3x}{(x^{2} + 6)^{2}} = \dfrac{Ax+B}{x^2+6} + \dfrac{Cx+D}{(x^2+6)^2} {/eq}

Multiplying both sides of the equation by the factored denominator, we have

{eq}x^3 - 3x = (Ax+B)(x^2+6) + Cx+D\\ x^3 - 3x = Ax^3 + Bx^2 + (6A+C)x + (6B+D)\\ {/eq}

Comparing coefficients, we have

{eq}1=A\\ 0=B\\ -3 = 6A+C\\ 0 = 6B+D\\ {/eq}

Solving this system, we have

{eq}A=1, B=0, C=-9, D=0 {/eq}

So the partial fraction decomposition is

{eq}\dfrac{x^{3} - 3x}{(x^{2} + 6)^{2}} = \dfrac{x}{x^2+6} + \dfrac{-9x}{(x^2+6)^2} {/eq}

Rewriting the integral,

{eq}\displaystyle \int \left(\dfrac{x}{x^2+6} + \dfrac{-9x}{(x^2+6)^2}\right) dx = \int \dfrac{x}{x^2+6}dx + \int\dfrac{-9x}{(x^2+6)^2}dx {/eq}

Using substitution for both integrals with {eq}u=x^2+6\\ du=2xdx {/eq} we have

{eq}\displaystyle\dfrac{1}{2}\int \dfrac{1}{u}du - \dfrac{9}{2}\int u^{-2}du = \dfrac{1}{2}\ln|u| +\dfrac{9}{2u} + c = \dfrac{1}{2}\ln|x^2+6| + \dfrac{9}{2(x^2+6)} + c {/eq}