👉 Try now NerdPal! Our new math app on iOS and Android

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

Step-by-step Solution

Go!
Math mode
Text mode
Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
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

Final Answer

$\frac{3}{2\left(x-2\right)^{2}}+\frac{-2}{\left(x-2\right)^{2}}+\frac{-4}{x-2}+\ln\left(x-2\right)+\frac{9}{x-2}+C_0$
Got another answer? Verify it here!

Step-by-step Solution

Specify the solving method

We could not solve this problem by using the method: Integrate by trigonometric substitution

1

We can factor the polynomial $x^3-6x^2+12x-8$ 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 $-8$

$1, 2, 4, 8$
2

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

$1$
3

The possible roots $\pm\frac{p}{q}$ of the polynomial $x^3-6x^2+12x-8$ will then be

$\pm1,\:\pm2,\:\pm4,\:\pm8$
4

Trying all possible roots, we found that $2$ is a root of the polynomial. When we evaluate it in the polynomial, it gives us $0$ as a result

$2^3-6\cdot 2^2+12\cdot 2-8=0$
5

Now, divide the polynomial by the root we found $\left(x-2\right)$ 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 $2$. Add the result to the second coefficient and then multiply this by $2$ and so on

$\left|\begin{array}{c}1 & -6 & 12 & -8 \\ & 2 & -8 & 8 \\ 1 & -4 & 4 & 0\end{array}\right|2$
6

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 $\left(x-2\right)$

$\int\frac{x^2-9x+15}{\left(x^{2}-4x+4\right)\left(x-2\right)}dx$
7

The trinomial $\left(x^{2}-4x+4\right)$ is a perfect square trinomial, because it's discriminant is equal to zero

$\Delta=b^2-4ac=-4^2-4\left(1\right)\left(4\right) = 0$
8

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{x^{2}}\:and\:b=\sqrt{4}$
9

Factoring the perfect square trinomial

$\int\frac{x^2-9x+15}{\left(x-2\right)^{2}\left(x-2\right)}dx$
10

When multiplying exponents with same base you can add the exponents: $\left(x-2\right)^{2}\left(x-2\right)$

$\int\frac{x^2-9x+15}{\left(x-2\right)^{3}}dx$
11

We can solve the integral $\int\frac{x^2-9x+15}{\left(x-2\right)^{3}}dx$ 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

$u=x-2$
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

Rewriting $x$ in terms of $u$

$x=u+2$
14

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

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

Expand the fraction $\frac{-3+\left(u+2\right)^2-9u}{u^{3}}$ into $3$ simpler fractions with common denominator $u^{3}$

$\int\left(\frac{-3}{u^{3}}+\frac{\left(u+2\right)^2}{u^{3}}+\frac{-9u}{u^{3}}\right)du$
16

Simplify the resulting fractions

$\int\left(\frac{-3}{u^{3}}+\frac{\left(u+2\right)^2}{u^{3}}+\frac{-9}{u^{2}}\right)du$
17

Expand the integral $\int\left(\frac{-3}{u^{3}}+\frac{\left(u+2\right)^2}{u^{3}}+\frac{-9}{u^{2}}\right)du$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{-3}{u^{3}}du+\int\frac{\left(u+2\right)^2}{u^{3}}du+\int\frac{-9}{u^{2}}du$
18

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

$\frac{3}{2\left(x-2\right)^{2}}$
19

The integral $\int\frac{\left(u+2\right)^2}{u^{3}}du$ results in: $\ln\left(x-2\right)+\frac{-4}{x-2}+\frac{-2}{\left(x-2\right)^{2}}$

$\ln\left(x-2\right)+\frac{-4}{x-2}+\frac{-2}{\left(x-2\right)^{2}}$
20

Gather the results of all integrals

$\frac{3}{2\left(x-2\right)^{2}}+\frac{-2}{\left(x-2\right)^{2}}+\frac{-4}{x-2}+\ln\left(x-2\right)+\int\frac{-9}{u^{2}}du$
21

The integral $\int\frac{-9}{u^{2}}du$ results in: $\frac{9}{x-2}$

$\frac{9}{x-2}$
22

Gather the results of all integrals

$\frac{3}{2\left(x-2\right)^{2}}+\frac{-2}{\left(x-2\right)^{2}}+\frac{-4}{x-2}+\ln\left(x-2\right)+\frac{9}{x-2}$
23

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

$\frac{3}{2\left(x-2\right)^{2}}+\frac{-2}{\left(x-2\right)^{2}}+\frac{-4}{x-2}+\ln\left(x-2\right)+\frac{9}{x-2}+C_0$

Final Answer

$\frac{3}{2\left(x-2\right)^{2}}+\frac{-2}{\left(x-2\right)^{2}}+\frac{-4}{x-2}+\ln\left(x-2\right)+\frac{9}{x-2}+C_0$

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 partial fractionsSolve 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 parts

Give us your feedback!

Function Plot

Plotting: $\frac{3}{2\left(x-2\right)^{2}}+\frac{-2}{\left(x-2\right)^{2}}+\frac{-4}{x-2}+\ln\left(x-2\right)+\frac{9}{x-2}+C_0$

SnapXam A2
Answer Assistant

beta
Got a different answer? Verify it!

Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
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: Quadratic Equations

The quadratic equations (or second degree equations) are those equations where the greatest exponent to which the unknown is raised is the exponent 2.

Your Math & Physics Tutor. Powered by AI

Available 24/7, 365.

Unlimited step-by-step math solutions. No ads.

Includes multiple solving methods.

Support for more than 100 math topics.

Premium access on our iOS and Android apps as well.

20% discount on online tutoring.

Choose your subscription plan:
Have a promo code?
Pay $39.97 USD securely with your payment method.
Please hold while your payment is being processed.
Create an Account