Partial Fractions with an irreducible quadratic factor

Question

$\int \frac{2}{(x-4)(x^2+2x+6)} dx$.

this is a partial fraction with irreducible quadratic factors. I know how to set it up and I found A, B, and C.

2 = A((x^2)+2x+6) +(x-4)(Bx+C). then I plugged in x = 4 to find A. to find the other values I distributed and equated coefficients.

$A = 1/15 \\ B = -1/15 \\ C = -6/15$

I am having trouble with integrating after I plug in the values. Any help is appreciated.

Answer 1

HINT:

Write as

$$\frac2{(x-4)(x^2+2x+6)}=\frac A{x-4}+\frac{B\frac{d(x^2+2x+6)}{dx}+C}{x^2+2x+6}$$

so that $$\int\frac{B\frac{d(x^2+2x+6)}{dx}}{x^2+2x+6}\ dx=B\ln|x^2+2x+6|+C$$

For $\displaystyle\int\frac C{x^2+2x+6}\ dx=\int\frac C{(x+1)^2+(\sqrt5)^2}\ dx$

set $x+1=\sqrt5\tan\theta$

Answer 2

So, you want to calculate:

$\dfrac{1}{15}\displaystyle{\int}\frac{dx}{x-4}-\dfrac{1}{15}\displaystyle{\int}\frac{x+6}{x^2+2x+6}dx$.

That first integral is just $\ln|x-4|$. For the second one, you want to separate it into two integrals: one natural log and one inverse tangent. Note that the derivative of the denominator is $2x+2=2(x+1)$. You want to see that $x+1$ factor in your numerator, so you rewrite:

$\begin{align} \displaystyle{\int}\frac{x+6}{x^2+2x+6}dx &= \displaystyle{\int}\frac{x+1+5}{x^2+2x+6}dx \\ &=\displaystyle{\int}\frac{x+1}{x^2+2x+6}dx + \displaystyle{\int}\frac{5}{x^2+2x+6}dx \\ &=\frac12\displaystyle{\int}\frac{2x+2}{x^2+2x+6}dx + \displaystyle{\int}\frac{5}{(x+1)^2+5}dx. \end{align}$

Note how that last step sets up the integral on the left for substitution with $u=x^2+2x+6$, and sets up the integral on the right for an inverse tangent antiderivative after using the substitution $u=x+1$.