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$$\int \frac{x^2-1}{x^4+x^3+x^2+x+1} dx$$

My attempt:

$$x^4+x^3+x^2+x+1=[2x^2+(1+\sqrt{5})x+2][2x^2+(1-\sqrt{5})x+2]$$

$$ x^4+x^3+x^2+x+1= \frac{ \frac{-4\sqrt{5}}{5}\cdot x -(\frac{\sqrt{5}}{5}+1) }{2x^2+(1+\sqrt{5})x+2}+ \frac{ \frac{4\sqrt{5}}{5}\cdot x +\frac{\sqrt{5}}{5}-1 }{2x^2+(1-\sqrt{5})x+2}$$

$$\int \frac{x^2-1}{x^4+x^3+x^2+x+1} dx= \int \left[\frac{ \frac{-4\sqrt{5}}{5}\cdot x -(\frac{\sqrt{5}}{5}+1) }{2x^2+(1+\sqrt{5})x+2}+ \frac{ \frac{4\sqrt{5}}{5}\cdot x +\frac{\sqrt{5}}{5}-1 }{2x^2+(1-\sqrt{5})x+2}\right]dx$$

When I try to solve the question after this part, I get the wrong answer every time. I would appreciate it if you could show me how to solve the next part.

Answer:$\frac{1}{\sqrt{5}}\cdot ln \left| \frac{2x^2+(1-\sqrt{5})x+2}{2x^2+(1+\sqrt{5})x+2} \right| + C$

Rehman
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    You can keep going with partial fractions. Notice $(x-1)(x^4+x^3+x^2+x+1)=x^5-1$ so the denominator factorizes over the non-real fifth roots of $1$. – user170231 Apr 21 '23 at 16:48

3 Answers3

3

Divide numerator and denominator by $x^2$

$$\int\dfrac{1-\frac{1}{x^2}}{x^2+x+\frac{1}{x}+\frac{1}{x^2}+1}dx$$

substitute $x+\frac{1}{x}=t$

So $(1-\frac{1}{x^2})dx = dt$

and $$x^2+\frac{1}{x^2} + 2=t^2$$

so

$$\int \frac{dt}{t^2-2+t+1}$$

Now proceed from here

Shlok Jain
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2

Another straightforward approach would be to divide numerator and denominator by $x^2$

$$\int \frac{1-\frac{1}{x^2}}{x^2+x+1+\frac{1}{x}+\frac{1}{x^2}}dx = \int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2+\left(x+\frac{1}{x}\right)-1}$$

And then complete the square

$$\int \frac{d\left(x+\frac{1}{x}+\frac{1}{2}\right)}{\left[\left(x+\frac{1}{x}\right)+\frac{1}{2}\right]^2-\frac{5}{4}}$$

which suggests using the substitution

$$x+\frac{1}{x} + \frac{1}{2} = \frac{\sqrt{5}}{2}\operatorname{coth} t\implies d\left(x+\frac{1}{2}+\frac{1}{2}\right) = - \frac{\sqrt{5}}{2} \operatorname{csch}^2 t \: dt$$

This simplifies the integrand to

$$\int -\frac{2}{\sqrt{5}} \:dt$$

and a final answer of

$$-\frac{2}{\sqrt{5}}\operatorname{coth}^{-1}\left[\frac{2}{\sqrt{5}}\left(x+\frac{1}{x}+\frac{1}{2}\right)\right]+C$$

which is equivalent to your given answer as inverse hyperbolic can be written as logs via Pythagoras.

Ninad Munshi
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2

Substitute $x=\frac{1-y}{1+y}$ to eliminate the first- and third-degree terms in the denominator, and to open the way for another substitution of $z=y^2+5$.

$$\begin{align*} &\int \frac{x^2-1}{x^4+x^3+x^2+x+1} \, dx \\ &= \int \frac{\frac{(1-y)^2}{(1+y)^2}-1}{\frac{(1-y)^4}{(1+y)^4}+\frac{(1-y)^3}{(1+y)^3}+\frac{(1-y)^2}{(1+y)^2}+\frac{(1-y)^4}{(1+y)^4}+1} \, \left(-\frac{2}{(1+y)^2}\right) \, dy \\ &= \int \frac{8y}{y^4+10y^2+5} \, dy \\ &= \int \frac{8y}{\left(y^2+5\right)^2-20} \, dy \\ &= \int \frac{4}{z^2-20} \, dz \end{align*}$$

user170231
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