Calculate $\int\frac{2x^5-5}{x^4-5x^2+6}dx$ with partial fractions

Question

I'm trying to calculate the following integral $$\int\frac{2x^5-5}{x^4-5x^2+6}dx$$

What I did was to divide polynomes: $(2x^5-5):( x^4-5x^2+6)=2x+\frac{10x^3-12x-5}{x^4-5x^2+6}$.

Then I have $$\int2x dx + \int\frac{10x^3-12x-5}{x^4-5x^2+6}dx$$

I used partial fractions on the second integral and I got $$\frac{Ax+B}{x^2-2}+\frac{Cx+D}{x^2-3}=\frac{10x^3-12x-5}{(x^2-2)(x^2-3)}$$

$$\Leftrightarrow (Ax+B)*(x^2-3)+(Cx+D)*(x^2-2)=10x^3-12x-5$$

However, I don't know how to continue. Are my steps so far correct?

Thanks

Answer 1

Actually, you should write $\dfrac{10x^3-12x+5}{x^4-5x^2+6}$ as$$\frac A{x+\sqrt 2}+\frac B{x-\sqrt2}+\frac C{x+\sqrt 3}+\frac D{x-\sqrt3}.$$If you write this expression as a single fraction, you will get\begin{multline}-A x^3+\sqrt{2} A x^2+3 A x-3 \sqrt{2} A-B x^3-\sqrt{2} B x^2+3 B x+3 \sqrt{2} B+\\-C x^3+\sqrt{3} C x^2+2 C x-2 \sqrt{3} C-D x^3-\sqrt{3} D x^2+2 D x+2 \sqrt{3} D\end{multline}divided by $x^4-5x^2+6$. So, solve the system$$\left\{\begin{array}{l}-3 \sqrt{2} A+3 \sqrt{2} B-2 \sqrt{3} C+2 \sqrt{3} D=5\\3 A+3 B+2 C+2 D=-12\\\sqrt{2} A-\sqrt{2} B+\sqrt{3} C-\sqrt{3} D=0\\-A-B-C-D=10.\end{array}\right.$$You will get that $A=4+\frac{5}{2 \sqrt{2}}$, $B=4-\frac{5}{2 \sqrt{2}}$, $C=-9-\frac{5}{2 \sqrt{3}}$, and $D=-9+\frac{5}{2 \sqrt{3}}$. Can you take it from here?

Answer 2

Since your equation needs to be an identity, just simply plug in $x=\pm\sqrt{2}$ to obtain $A$ and $B$.

$$\begin{cases}-(\sqrt{2}A+B)=8\sqrt{2}-5 \\ -(-\sqrt{2}A+B)=-8\sqrt{2}-5\end{cases}$$

It is then obvious that $A=-8,~B=5.$

Similarly plug in $x=\pm\sqrt{3}$ to obtain $C$ and $D.$

$$\begin{cases}\sqrt{3}C+D=18\sqrt{3}-5 \\ -\sqrt{3}C+D=-18\sqrt{3}-5\end{cases}$$

Again, obvious that $C=18,~D=-5.$

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By the way, if you're clever, there's no need to really plug in 4 values. Just plug in $x=\sqrt{2}$ and $x=\sqrt{3},$ since you already know that $A,~B,~C,~D$ are rational.

Answer 3

First, rewrite,

$$I=\int\frac{2x^5-5}{x^4-5x^2+6}dx =\int\frac{t^2}{t^2-5t+6}dt-\int\frac{5}{(x^2-2)(x^2-3)}dx$$

where $t=x^2$ is used to simplify the first integral. Then decompose,

$$I=x^2 + \int\frac{9}{t-3}dt-\int\frac{4}{t-2}dt -\int\frac{5}{x^2-3}dx+\int\frac{5}{x^2-2}dx$$