Find the integral $\int\frac{x+8}{x^6-2x^4+x^2}dx$

We can factor the polynomial $x^6-2x^4+x^2$ using the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $0$

Next, list all divisors of the leading coefficient $a_n$, which equals $1$

The possible roots $\pm\frac{p}{q}$ of the polynomial $x^6-2x^4+x^2$ will then be

We can factor the polynomial $x^6-2x^4+x^2$ using synthetic division (Ruffini's rule). We found that $1$ is a root of the polynomial

Now, divide the polynomial by the root we found $\left(x-1\right)$ using synthetic division (Ruffini's rule). First, write the coefficients of the terms of the numerator in descending order. Then, take the first coefficient $1$ and multiply by the factor $1$. Add the result to the second coefficient and then multiply this by $1$ and so on

In the last row of the division appear the new coefficients, with remainder equals zero. Now, rewrite the polynomial (a degree less) with the new coefficients, and multiplied by the factor $\left(x-1\right)$

We can factor the polynomial $\left(x^{5}+x^{4}-x^{3}-x^{2}\right)$ using the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $0$

Next, list all divisors of the leading coefficient $a_n$, which equals $1$

The possible roots $\pm\frac{p}{q}$ of the polynomial $\left(x^{5}+x^{4}-x^{3}-x^{2}\right)$ will then be

We can factor the polynomial $\left(x^{5}+x^{4}-x^{3}-x^{2}\right)$ using synthetic division (Ruffini's rule). We found that $1$ is a root of the polynomial

Now, divide the polynomial by the root we found $\left(x-1\right)$ using synthetic division (Ruffini's rule). First, write the coefficients of the terms of the numerator in descending order. Then, take the first coefficient $1$ and multiply by the factor $1$. Add the result to the second coefficient and then multiply this by $1$ and so on

In the last row of the division appear the new coefficients, with remainder equals zero. Now, rewrite the polynomial (a degree less) with the new coefficients, and multiplied by the factor $\left(x-1\right)$

We can factor the polynomial $\left(x^{4}+2x^{3}+x^{2}\right)$ using the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $0$

Next, list all divisors of the leading coefficient $a_n$, which equals $1$

The possible roots $\pm\frac{p}{q}$ of the polynomial $\left(x^{4}+2x^{3}+x^{2}\right)$ will then be

We can factor the polynomial $\left(x^{4}+2x^{3}+x^{2}\right)$ using synthetic division (Ruffini's rule). We found that $-1$ is a root of the polynomial

Now, divide the polynomial by the root we found $\left(x+1\right)$ using synthetic division (Ruffini's rule). First, write the coefficients of the terms of the numerator in descending order. Then, take the first coefficient $1$ and multiply by the factor $-1$. Add the result to the second coefficient and then multiply this by $-1$ and so on

In the last row of the division appear the new coefficients, with remainder equals zero. Now, rewrite the polynomial (a degree less) with the new coefficients, and multiplied by the factor $\left(x+1\right)$

Factor the polynomial $\left(x^{3}+x^{2}\right)$ by it's greatest common factor (GCF): $x^2$

Rewrite the fraction $\frac{x+8}{\left(x-1\right)^2\left(x+1\right)^2x^2}$ in $6$ simpler fractions using partial fraction decomposition

Find the values for the unknown coefficients: $A, B, C, D, F, G$. The first step is to multiply both sides of the equation from the previous step by $\left(x-1\right)^2\left(x+1\right)^2x^2$

Multiplying polynomials

Simplifying

Assigning values to $x$ we obtain the following system of equations

Proceed to solve the system of linear equations

Rewrite as a coefficient matrix

Reducing the original matrix to a identity matrix using Gaussian Elimination

The integral of $\frac{x+8}{\left(x-1\right)^2\left(x+1\right)^2x^2}$ in decomposed fraction equals

Expand the integral $\int\left(\frac{9}{4\left(x-1\right)^2}+\frac{7}{4\left(x+1\right)^2}+\frac{32}{37x^2}+\frac{-2.943243}{x-1}+\frac{2.714414}{x+1}+\frac{\frac{49}{370}}{x}\right)dx$ into $6$ integrals using the sum rule for integrals, to then solve each integral separately

We can solve the integral $\int\frac{9}{4\left(x-1\right)^2}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $x-1$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

Gather the results of all integrals

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$