Integral Calculus Question: $ g\left(x\right)=\int\frac{x^2-12}{\left(x^2-6x+k^2\right)^2}dx $ where $k \in \Bbb N$

Question

Assume that $$ g\left(x\right)=\int\frac{x^2-12}{\left(x^2-6x+k\right)^2}dx, $$ where $k \in \Bbb N$, is a rational function. Find sum of all possible values of k.

My attempt: \begin{align*} g\left(x\right) &=\int\frac{x^2-12}{\left(x^2-6x+k\right)^2}dx \\ &=\int\frac{1}{x^2-6x+k}dx + 3\int\frac{\left(2x-6\right)}{\left(x^2-6k+k\right)^2}dx + \left(6-k\right)\int\frac{1}{\left(x^2-6k+k\right)^2}dx \\ &=\int\frac{1}{\left(x-3\right)^2+k-9}dx + 3\int\frac{\left(2x-6\right)}{\left(x^2-6k+k\right)^2}dx + \left(6-k\right)\int\frac{1}{\left(x^2-6k+k\right)^2}dx \end{align*}

I know $k=9$ is a solution as it makes the quadratic in the denominator a perfect square but how am I supposed to go about finding the other solutions? Is there a better approach than using partial fractions? I also don't know how to integrate the last term (dx/biquadratic)...

Answer 1

You can make it faster writing $$P=\frac{x^2-12}{\left(x^2-6 x+k\right)^2}=\frac{x^2-12}{(x-a)^2\,(x-b)^2}$$

Using partial fraction decomposition $$P=\frac{2 (a b-12)}{(a-b)^3}\left(\frac 1{x-b}-\frac 1{x-a}\right)+\frac 1{(a-b)^2}\left(\frac{a^2-12}{(x-a)^2}+\frac{b^2-12}{(x-b)^2} \right)$$

Integrate term wise and look at the limit of the result when $b\to a$.

Answer 2

Let $g(x)=\tfrac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials with real coefficients and $p(x)=\sum_{i=0}^{n} a_ix^i$. Dividing, if needed, both the numerator and the denominator of $g$ by their greatest common divisor, we can assume that $p(x)$ and $q(x)$ have no common roots and the highest coefficient of $q(x)$ is $1$. Then $$\frac{x^2-12}{(x^2-6x+k)^2}=g'(x)=\frac{p'(x)q(x)-q'(x)p(x)}{q(x)^2}.$$ Since $k$ is natural, $\pm\sqrt{12}$ is not a root of the polynomial $x^2-6x+k$, so the fraction $\tfrac{x^2-12}{(x^2-6x+k)^2}$ is irreducible. Let $r(x)=\gcd (q(x),q'(x))$ be the polynomial with the highest coefficient $1$. Note that $\deg r(x)=\deg q(x)-R$, where $R$ is the number of distinct roots of $q(x)$. The irreducible fraction for $\tfrac{p'(x)q(x)-q'(x)p(x)}{q(x)^2}$ has the denominator $\tfrac {q(x)^2}{r(x)}$. Thus $$4=2\deg q(x)-\deg r(x)=\deg q(x)+R.$$ So the following cases are possible.

1)) $\deg q(x)=2$. Then $q(x)=x^2-6x+k$. Then $$x^2-12=p'(x)q(x)-q'(x)p(x)=p'(x)(x^2-6x+k)-(2x-6)p(x).$$ If $n\ge 3$ then the highest coefficient of $p'(x)(x^2-6x+k)-(2x-6)p(x)$ is $(n-2)a_n\ne 0$ at $x^{n+1}$, so $n\le 2$. Then $$x^2-12=(2a_2x+a_1)(x^2-6x+k)-(2x-6)(a_2x^2+a_1x+a_0)=$$ $$(-a_1-6a_2)x^2+2(ka_2-a_0)x+(ka_1+6a_0).$$ So $-1=a_1+6a_2$, $0=ka_2-a_0$, and $-12=ka_1+6a_0$. Thus $12=-(ka_1+6a_0)=-(ka_1+6ka_2)=k$.

2)) $\deg q(x)=3$. Then $R=1$, so the polynomial $x^2-6x+k$ has the unique root, which we denote by $x_0$. By Vieta's formulae, $2x_0=6$ and $x_0^2=9$, so $x_0=3$ and $k=9$. Then $q(x)=(x-3)^3$ and $$\frac{x^2-12}{(x-3)^4}=g'(x)=\frac{p'(x)q(x)-q'(x)p(x)}{q(x)^2}=$$ $$\frac{p'(x)(x-3)^3-3(x-3)^2p(x)}{(x-3)^6}=$$ $$\frac{p'(x)(x-3)-3p(x)}{(x-3)^4}.$$ Thus $p'(x)(x-3)-3p(x)=x^2-12$. If $n\ge 4$ then the highest coefficient of $p'(x)(x-3)-3p(x)$ is $(n-3)a_n\ne 0$ at $x^n$, so $n\le 3$. Then $x^2-12=(3a_3x^2+2a_2x+a_1)(x-3)-3(a_3x^3+a_2x^2+a_1x+a_0)$, so $1=-9a_3-a_2$, $0=-6a_2-2a_1$, and $-12=-3a_1-3a_0$.

Answer 3

We want to find $k$ such that there are real coefficient polynomials $f(x),g(x)$ satisfying $$\frac{x^2-12}{(x^2-6x+k)^2}=\bigg(\frac{g(x)}{f(x)}\bigg)'=\frac{g'(x)f(x)-g(x)f'(x)}{(f(x))^2}\tag1$$ and $\gcd(f(x),g(x))=1$ with $f(x)\not=0$.

It follows from $(1)$ that $$(x^2-12)(f(x))^2=(x^2-6x+k)^2(g'(x)f(x)-g(x)f'(x))\tag2$$ which implies $x^2-6x+k\mid (x^2-12)(f(x))^2$.

Suppose that $x^2-6x+k$ and $x^2-12$ have a common root $x=a$. Then, we have $a^2-6a+k=0$ and $a^2-12=0$. These give $k=-12\pm 12\sqrt 3$ which is impossible. Therefore, $\gcd(x^2-6x+k,x^2-12)=1$.

So, $x^2-6x+k\mid f(x)$, and we can write $f(x)=(x^2-6x+k)h(x)$ where $h(x)$ is a real coefficient polynomial.

Then, from $(2)$, $$(x^2-12)(h(x))^2=g'(x)(x^2-6x+k)h(x)-g(x)\bigg((2x-6)h(x)+(x^2-6x+k)h'(x)\bigg)$$ i.e. $$h(x)\bigg((x^2-12)h(x)-g'(x)(x^2-6x+k)+(2x-6)g(x)\bigg)=-(x^2-6x+k)h'(x)g(x)\tag3$$

Since $\gcd(h(x),g(x))=1$, it follows that $h(x)\mid (x^2-6x+k)h'(x)$.

Let us first consider the case where $h(x)$ is a non-zero constant function $C$. Then, $(3)$ implies that $$(x^2-12)C-g'(x)(x^2-6x+k)+(2x-6)g(x)=0$$ We write $g(x)=\displaystyle\sum_{j=0}^{m}a_jx^j$ with $a_m\not=0$. If $m\ge 3$, then the coefficient of $x^{m+1}$ has to be $(2-m)a_m=0$ which is impossible. So, $m\le 2$, and we have $C(x^2-12)-(2a_2x+a_1)(x^2-6x+k)+(2x-6)(a_2x^2+a_1x+a_0)=0$. Solving $C+12a_2-a_1+2a_1-6a_2=0,-2a_2k+6a_1+2a_0-6a_1=0$ and $-12C-a_1k-6a_0=0$ gives $k=12$ which is sufficient since $\frac{x^2-12}{(x^2-6x+12)^2}=\bigg(\frac{-x}{x^2-6x+12}\bigg)'$.

Next, let us consider the case where $h(x)$ is not a constant function.

• For $0\le k\le 8$, $x^2-6x+k=0$ has two distinct real roots $a=3-\sqrt{9-k},b=3+\sqrt{9-k}$. Since $h(x)\mid (x^2-6x+k)h'(x)$, we can write $h(x)=(x-a)^c(x-b)^dj(x)$ where $c,d$ are non-negative integers with $(c,d)\not=(0,0)$ and $\gcd(j(x),x^2-6x+k)=1$. If $c\ge 1$ and $d\ge 1$, since $$\begin{align}&\frac{(x^2-6x+k)h'(x)}{h(x)} \\\\&=\frac{c(x-a)^{c}(x-b)^{d+1}j(x)+d(x-a)^{c+1}(x-b)^{d}j(x)+(x-a)^{c+1}(x-b)^{d+1}j'(x)}{h(x)} \\\\&=c(x-b)+d(x-a)+\frac{(x^2-6x+k)j'(x)}{j(x)}\end{align}$$ we have to have $j(x)\mid (x^2-6x+k)j'(x)$. Since $\gcd(j(x),x^2-6x+k)=1$, $j(x)$ has to be a non-zero constant function with $j'(x)=0$, so $(3)$ implies $$(x^2-6x+k)\bigg((x^2-12)(x-a)^{c-1}(x-b)^{d-1}j(x)-g'(x)\bigg)=\bigg((-c-d-2)x+bc+ad+6\bigg)g(x)$$ Since $x^2-6x+k\mid (-c-d-2)x+bc+ad+6$, we have to have $-c-d-2=0$, i.e. $c+d=-2$ (and $bc+ad+6=0$) which is impossible. So, we have to have either $c=0$ or $d=0$. If $c=0$, then $(3)$ implies $$2(x-3)g(x)=(x-a)\bigg(-dg(x)-(x^2-12)(x-b)+g'(x)(x-b)\bigg)$$ Since $\gcd(x-a,g(x))=1$, $x-a\mid 2(x-3)$ which is impossible. Similarly, if $d=0$, we have to have $x-b\mid 2(x-3)$ which is impossible.

• $k=9$ is sufficient since $\frac{x^2-12}{(x^2-6x+9)^2}=\bigg(\frac{-x^2+3x+1}{(x-3)^3}\bigg)'$.

• For $k\gt 9$, $x^2-6x+9$ has two distinct non-real roots. Since $h(x)\mid (x^2-6x+k)h'(x)$, we can write $h(x)=(x-a)^c(x-b)^dp(x)$ where $c,d$ are non-negative integers with $(c,d)\not=(0,0)$ and $\gcd(p(x),x^2-6x+k)=1$. Moreover, if $c\not=d$, then $h(x)$ has at least one non-real coefficient. So, we have to have $c=d$, and so we can write $h(x)=(x^2-6x+k)^rp(x)$ where $r$ is a positive integer. Since $h(x)\mid (x^2-6x+k)h'(x)$, it follows from $\frac{(x^2-6x+k)h'(x)}{h(x)}=r+\frac{(x^2-6x+k)p'(x)}{p(x)}$ that $p(x)$ has to be a non-zero constant function $C$. Then, $(3)$ implies $$(x^2-6x+k)\bigg((x^2-12)(x^2-6x+k)^{r-1}C-g'(x)\bigg)=(-2x-r+6)g(x)$$ Since $\gcd(x^2-6x+k,g(x))=1$, we have to have $x^2-6x+k\mid -2x-r+6$ which is impossible.

Since $k=9,12$ are the only possible values of $k$, the sum of them is $\color{red}{21}$.

Answer 4

$$ g(x)=\dfrac {(k-12)\cdot\arctan\left(\dfrac{x-3}{\sqrt{k-9}}\right) }{2(k-9)^{3/2}}+\frac{x(6-k)-3(k-12)}{2(k-9)(x^{2}-6x+k)} $$