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Find the integral $\int\frac{4x-1}{x^3-2x^2-x}dx$

Step-by-step Solution

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Final answer to the problem

$\ln\left(x\right)-\frac{1}{2}\ln\left(-2+\left(x-1\right)^2\right)-\frac{5\sqrt{2}}{2}\ln\left(\frac{0.4142136+x}{\sqrt{-2+\left(x-1\right)^2}}\right)+C_0$
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Step-by-step Solution

Specify the solving method

1

Rewrite the expression $\frac{4x-1}{x^3-2x^2-x}$ inside the integral in factored form

$\int\frac{4x-1}{x\left(x^2-2x-1\right)}dx$
2

Rewrite the fraction $\frac{4x-1}{x\left(x^2-2x-1\right)}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{4x-1}{x\left(x^2-2x-1\right)}=\frac{A}{x}+\frac{Bx+C}{x^2-2x-1}$
3

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 $x\left(x^2-2x-1\right)$

$4x-1=x\left(x^2-2x-1\right)\left(\frac{A}{x}+\frac{Bx+C}{x^2-2x-1}\right)$
4

Multiplying polynomials

$4x-1=\frac{x\left(x^2-2x-1\right)A}{x}+\frac{x\left(x^2-2x-1\right)\left(Bx+C\right)}{x^2-2x-1}$
5

Simplifying

$4x-1=\left(x^2-2x-1\right)A+x\left(Bx+C\right)$
6

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

$\begin{matrix}-1=-A&\:\:\:\:\:\:\:(x=0) \\ 3=-2A+B+C&\:\:\:\:\:\:\:(x=1) \\ -5=2A+B-C&\:\:\:\:\:\:\:(x=-1)\end{matrix}$
7

Proceed to solve the system of linear equations

$\begin{matrix} -1A & + & 0B & + & 0C & =-1 \\ -2A & + & 1B & + & 1C & =3 \\ 2A & + & 1B & - & 1C & =-5\end{matrix}$
8

Rewrite as a coefficient matrix

$\left(\begin{matrix}-1 & 0 & 0 & -1 \\ -2 & 1 & 1 & 3 \\ 2 & 1 & -1 & -5\end{matrix}\right)$
9

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 6\end{matrix}\right)$
10

The integral of $\frac{4x-1}{x\left(x^2-2x-1\right)}$ in decomposed fraction equals

$\int\left(\frac{1}{x}+\frac{-x+6}{x^2-2x-1}\right)dx$
11

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

$\int\frac{1}{x}dx+\int\frac{-x+6}{x^2-2x-1}dx$
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

$\ln\left(x\right)+\int\frac{-x+6}{x^2-2x-1}dx$
14

Rewrite the expression $\frac{-x+6}{x^2-2x-1}$ inside the integral in factored form

$\ln\left(x\right)+\int\frac{-x+6}{-2+\left(x-1\right)^2}dx$
15

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

$u=x-1$
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

$du=dx$
17

Rewriting $x$ in terms of $u$

$x=u+1$
18

Substituting $u$, $dx$ and $x$ in the integral and simplify

$\ln\left(x\right)+\int\frac{5-u}{-2+u^2}du$
19

The integral $\int\frac{5-u}{-2+u^2}du$ results in: $-\frac{5\sqrt{2}}{2}\ln\left(\frac{x-1}{\sqrt{-2+\left(x-1\right)^2}}+\frac{\sqrt{2}}{\sqrt{-2+\left(x-1\right)^2}}\right)-\frac{1}{2}\ln\left(-2+\left(x-1\right)^2\right)$

$-\frac{5\sqrt{2}}{2}\ln\left(\frac{x-1}{\sqrt{-2+\left(x-1\right)^2}}+\frac{\sqrt{2}}{\sqrt{-2+\left(x-1\right)^2}}\right)-\frac{1}{2}\ln\left(-2+\left(x-1\right)^2\right)$
20

Gather the results of all integrals

$\ln\left(x\right)-\frac{1}{2}\ln\left(-2+\left(x-1\right)^2\right)-\frac{5\sqrt{2}}{2}\ln\left(\frac{x-1}{\sqrt{-2+\left(x-1\right)^2}}+\frac{\sqrt{2}}{\sqrt{-2+\left(x-1\right)^2}}\right)$
21

The least common multiple (LCM) of a sum of algebraic fractions consists of the product of the common factors with the greatest exponent, and the uncommon factors

$L.C.M.=\sqrt{-2+\left(x-1\right)^2}$
22

Combine and simplify all terms in the same fraction with common denominator $\sqrt{-2+\left(x-1\right)^2}$

$\ln\left(x\right)-\frac{1}{2}\ln\left(-2+\left(x-1\right)^2\right)-\frac{5\sqrt{2}}{2}\ln\left(\frac{0.4142136+x}{\sqrt{-2+\left(x-1\right)^2}}\right)$
23

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

$\ln\left(x\right)-\frac{1}{2}\ln\left(-2+\left(x-1\right)^2\right)-\frac{5\sqrt{2}}{2}\ln\left(\frac{0.4142136+x}{\sqrt{-2+\left(x-1\right)^2}}\right)+C_0$

Final answer to the problem

$\ln\left(x\right)-\frac{1}{2}\ln\left(-2+\left(x-1\right)^2\right)-\frac{5\sqrt{2}}{2}\ln\left(\frac{0.4142136+x}{\sqrt{-2+\left(x-1\right)^2}}\right)+C_0$

Explore different ways to solve this problem

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

Solve integral of ((4x-1)/(x^3+-2x^2))dx using partial fractionsSolve integral of ((4x-1)/(x^3+-2x^2))dx using basic integralsSolve integral of ((4x-1)/(x^3+-2x^2))dx using u-substitutionSolve integral of ((4x-1)/(x^3+-2x^2))dx using integration by parts

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Function Plot

Plotting: $\ln\left(x\right)-\frac{1}{2}\ln\left(-2+\left(x-1\right)^2\right)-\frac{5\sqrt{2}}{2}\ln\left(\frac{0.4142136+x}{\sqrt{-2+\left(x-1\right)^2}}\right)+C_0$

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a
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u
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x
y
z
.
(◻)
+
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×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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