Find the values for the unknown coefficients: $A, B, C$. The first step is to multiply both sides of the equation from the previous step by $x\left(x^2-2x-1\right)$
Expand the integral $\int\left(\frac{1}{x}+\frac{-x+6}{x^2-2x-1}\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately
$\int\frac{1}{x}dx+\int\frac{-x+6}{x^2-2x-1}dx$
Intermediate steps
12
The integral $\int\frac{1}{x}dx$ results in: $\ln\left(x\right)$
$\ln\left(x\right)$
13
Gather the results of all integrals
$\ln\left(x\right)+\int\frac{-x+6}{x^2-2x-1}dx$
Intermediate steps
14
Rewrite the expression $\frac{-x+6}{x^2-2x-1}$ inside the integral in factored form
We can solve the integral $\int\frac{-x+6}{-2+\left(x-1\right)^2}dx$ 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
16
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
17
Rewriting $x$ in terms of $u$
$x=u+1$
Intermediate steps
18
Substituting $u$, $dx$ and $x$ in the integral and simplify
$\ln\left(x\right)+\int\frac{5-u}{-2+u^2}du$
Intermediate steps
19
The integral $\int\frac{5-u}{-2+u^2}du$ results in: $5\int\frac{1}{-2+u^2}du+\left(\frac{\sqrt{2}}{4}\ln\left(0.4142136+x\right)-\frac{\sqrt{2}}{4}\ln\left(-2.4142136+x\right)\right)\left(x-1\right)-\frac{\sqrt{2}}{4}\left(\left(0.4142136+x\right)\ln\left(0.4142136+x\right)-\left(x-1\right)-\sqrt{2}\right)+\frac{\sqrt{2}}{4}\left(\left(-2.4142136+x\right)\ln\left(-2.4142136+x\right)-\left(x-1\right)+\sqrt{2}\right)$
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.