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Find the integral $\int\frac{x^2-9x+15}{x^3-6x^2+12x-8}dx$

Step-by-step Solution

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Final Answer

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

Specify the solving method

1

Rewrite the expression $\frac{x^2-9x+15}{x^3-6x^2+12x-8}$ inside the integral in factored form

$\int\frac{x^2-9x+15}{\left(x-2\right)^{3}}dx$
2

Rewrite the fraction $\frac{x^2-9x+15}{\left(x-2\right)^{3}}$ in $3$ simpler fractions using partial fraction decomposition

$\frac{x^2-9x+15}{\left(x-2\right)^{3}}=\frac{A}{x-2}+\frac{B}{\left(x-2\right)^{2}}+\frac{C}{\left(x-2\right)^{3}}$
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 $\left(x-2\right)^{3}$

$x^2-9x+15=\left(x-2\right)^{3}\left(\frac{A}{x-2}+\frac{B}{\left(x-2\right)^{2}}+\frac{C}{\left(x-2\right)^{3}}\right)$
4

Multiplying polynomials

$x^2-9x+15=\frac{\left(x-2\right)^{3}A}{x-2}+\frac{\left(x-2\right)^{3}B}{\left(x-2\right)^{2}}+\frac{\left(x-2\right)^{3}C}{\left(x-2\right)^{3}}$
5

Simplifying

$x^2-9x+15=\left(x-2\right)^{2}A+\left(x-2\right)B+C$
6

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

$\begin{matrix}15=4A-2B+C&\:\:\:\:\:\:\:(x=0) \\ 7=A-B+C&\:\:\:\:\:\:\:(x=1) \\ 25=9A-3B+C&\:\:\:\:\:\:\:(x=-1)\end{matrix}$
7

Proceed to solve the system of linear equations

$\begin{matrix}4A & - & 2B & + & 1C & =15 \\ 1A & - & 1B & + & 1C & =7 \\ 9A & - & 3B & + & 1C & =25\end{matrix}$
8

Rewrite as a coefficient matrix

$\left(\begin{matrix}4 & -2 & 1 & 15 \\ 1 & -1 & 1 & 7 \\ 9 & -3 & 1 & 25\end{matrix}\right)$
9

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 1\end{matrix}\right)$
10

The integral of $\frac{x^2-9x+15}{\left(x-2\right)^{3}}$ in decomposed fraction equals

$\int\left(\frac{1}{x-2}+\frac{-5}{\left(x-2\right)^{2}}+\frac{1}{\left(x-2\right)^{3}}\right)dx$
11

Expand the integral $\int\left(\frac{1}{x-2}+\frac{-5}{\left(x-2\right)^{2}}+\frac{1}{\left(x-2\right)^{3}}\right)dx$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{1}{x-2}dx+\int\frac{-5}{\left(x-2\right)^{2}}dx+\int\frac{1}{\left(x-2\right)^{3}}dx$
12

We can solve the integral $\int\frac{1}{x-2}dx$ 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 $x-2$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=x-2$
13

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

$du=dx$
14

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

$\int\frac{1}{u}du+\int\frac{-5}{\left(x-2\right)^{2}}dx+\int\frac{1}{\left(x-2\right)^{3}}dx$
15

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

$\ln\left(x-2\right)$
16

Gather the results of all integrals

$\ln\left(x-2\right)+\int\frac{-5}{\left(x-2\right)^{2}}dx+\int\frac{1}{\left(x-2\right)^{3}}dx$
17

We can solve the integral $\int\frac{-5}{\left(x-2\right)^{2}}dx$ 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 $x-2$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=x-2$
18

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

$du=dx$
19

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

$\ln\left(x-2\right)+\int\frac{-5}{u^{2}}du+\int\frac{1}{\left(x-2\right)^{3}}dx$
20

The integral $\int\frac{-5}{u^{2}}du$ results in: $\frac{5}{x-2}$

$\frac{5}{x-2}$
21

Gather the results of all integrals

$\ln\left(x-2\right)+\frac{5}{x-2}+\int\frac{1}{\left(x-2\right)^{3}}dx$
22

We can solve the integral $\int\frac{1}{\left(x-2\right)^{3}}dx$ 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 $x-2$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=x-2$
23

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

$du=dx$
24

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

$\ln\left(x-2\right)+\frac{5}{x-2}+\int\frac{1}{u^{3}}du$
25

The integral $\int\frac{1}{u^{3}}du$ results in: $\frac{1}{-2\left(x-2\right)^{2}}$

$\frac{1}{-2\left(x-2\right)^{2}}$
26

Gather the results of all integrals

$\ln\left(x-2\right)+\frac{5}{x-2}+\frac{1}{-2\left(x-2\right)^{2}}$
27

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

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

Final Answer

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

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 ((x^2+-9x)/(x^3+-6x^2))dx using basic integralsSolve integral of ((x^2+-9x)/(x^3+-6x^2))dx using u-substitutionSolve integral of ((x^2+-9x)/(x^3+-6x^2))dx using integration by partsSolve integral of ((x^2+-9x)/(x^3+-6x^2))dx using trigonometric substitution

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Function Plot

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

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x
y
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.
(◻)
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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

How to improve your answer:

Main Topic: Integrals by Partial Fraction Expansion

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.

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