Find the values for the unknown coefficients: $A, B, C, D, F$. The first step is to multiply both sides of the equation from the previous step by $\left(x^2-1\right)^2\left(x-6\right)$
Expand the integral $\int\left(\frac{-\frac{11}{35}x+\frac{4}{35}}{\left(x^2-1\right)^2}+\frac{\frac{3}{334}}{x-6}+\frac{-\frac{3}{334}x-\frac{16}{297}}{x^2-1}\right)dx$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately
We can solve the integral $\int\frac{\frac{3}{334}}{x-6}dx$ 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 $x-6$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part
$u=x-6$
Intermediate steps
12
Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above
$du=dx$
13
Substituting $u$ and $dx$ in the integral and simplify
The integral $\int\frac{-\frac{11}{35}x+\frac{4}{35}}{\left(x^2-1\right)^2}dx$ results in: $\int\frac{-\frac{11}{35}x+\frac{4}{35}}{\left(x+1\right)^2\left(x-1\right)^2}dx$
The integral $\int\frac{\frac{3}{334}}{u}du$ results in: $\frac{3}{334}\ln\left(x-6\right)$
$\frac{3}{334}\ln\left(x-6\right)$
Intermediate steps
16
The integral $\int\frac{-\frac{3}{334}x-\frac{16}{297}}{x^2-1}dx$ results in: $-\frac{3}{334}\int\frac{x}{\left(x+1\right)\left(x-1\right)}dx+\frac{8}{297}\ln\left(\frac{1+x}{-1+x}\right)$
We can solve the integral $\int\frac{-\frac{11}{35}x+\frac{4}{35}}{\left(x+1\right)^2\left(x-1\right)^2}dx$ 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
$u=x+1$
Intermediate steps
19
Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above
$du=dx$
Intermediate steps
20
Rewriting $x$ in terms of $u$
$x=u-1$
Intermediate steps
21
Substituting $u$, $dx$ and $x$ in the integral and simplify
The integral $\int\frac{\frac{3}{7}-\frac{11}{35}u}{u^2\left(-2+u\right)^2}du$ results in: $\frac{-3}{28\left(x+1\right)}+\frac{1}{20\left(-2+x+1\right)}+\frac{3}{140}\ln\left(x+1\right)-\frac{1}{70}\ln\left(x+1-2\right)$
The integral $-\frac{3}{334}\int\frac{x}{\left(x+1\right)\left(x-1\right)}dx$ results in: $-\frac{1}{223}\ln\left(x+1\right)-\frac{1}{223}\ln\left(x-1\right)$
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more
The partial fraction decomposition or partial fraction expansion of a rational function is the operation that consists in expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.