$\int\frac{9x+9}{(x-1)(x^2+4x+13)}dx$

Question

Integral of $\frac {9x+9}{(x-1)(x^2+4x+13)}$

I am dumb founded by this question.

I know that the integral of $\frac {1}{x^2+4x+13}$ is equal to $\frac 13 \arctan(x+\frac {2}{3})+C$ If you set $dv= \frac {1}{x^2+4x+13}, v=\frac 13 \arctan(x+2/3)+C$ and $u=\frac {x+1}{x-1}, du= -\frac 2{(x-1)^2}$ integration by parts won't work.

I tried multiplying $(x-1)(x^2+4x+13)$ To get the bottom part which is $(x^3+3x^2+9x-13)$. Incidentally the derivative of $\ln(x^3+3x^2+9x-13)$ is $\frac {(x+2)(3x+3)}{x^3+3x^2+9x-13}.$ You can arrange $\int \frac {9x+9}{(x-1)(x^2+4x+13)}\ dx$ to $3\int \frac {(x+2)(3x+3)}{(x+2)(x^3+3x^2+9x-13)}\ dx$. But integration by parts won't work if you do that either.

Answer 1

As $x^2+4x+13$ is irreducible, the partial fraction decomposition theorem says you can write $$ \frac{9x+9}{(x-1)(x^2+4x+13)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+4x+13}.$$ To determine the coefficients, multiply both sides by $(x-1)(x^2+4x+13)$: $$9(x+1)=A(x^2+4x+13)+(Bx+C)(x-1),$$ and set successively in this equality:

• $x=1$, which yields $18=18 A+(B+C)\cdot 0$, whence $A=1$;
• $x=-2+3i$ (one of the complex roots of $x^2+4x+13$): \begin{align}9(-1+3i)&=A\cdot 0+\bigl(B(-2+3i)+C\bigr)(-3+3i)\\ \iff9(-1+3i)&=B-C+i(-3B+C)\\ \iff&\phantom{=}\begin{cases}\phantom{-3}B+C=39\\ C-5B=9 \end{cases}\iff B=-1,\enspace C=4. \end{align} So you finally have to calculate $$\int\!\frac{\mathrm d x}{x+1}-\int\!\frac{x\,\mathrm d x}{x^2+4x+13}+4\int\!\frac{\mathrm d x}{x^2+4x+13}.$$ Can you proceed?

Answer 2

Partial fractions.

$\frac {9x+9}{(x-1)(x^2+4x+13)} = \frac {A}{x-1} + \frac {Bx+C}{(x^2+4x+13)}\\ 9x+9 = A(x^2 + 4x + 13) + (Bx+C)(x-1)\\ 9x+9 = (A+B) x^2 + (4A-B+C)x + (13A - C)$

Solve this system of linear equations

$A+B = 0\\4A-B+C = 9\\13A-C = 9$

And apply to the partial fractions.

$\frac {9x+9}{(x-1)(x^2+4x+13)} = \frac {1}{x-1} + \frac {-x + 4}{(x^2+4x+13)}$

Now you want to do a $u$ substitution to take care of the right hand term.

$\frac {1}{x-1} - \frac {x + 2}{(x^2+4x+13)} + \frac {6}{(x^2+4x+13)}$