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Find the integral $\int\frac{3x+5}{x^3-x^2-x+1}dx$

Step-by-step Solution

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Final Answer

$\frac{-4}{x-1}+\frac{1}{2}\ln\left(1+x\right)-\frac{1}{2}\ln\left(x-1\right)+C_0$
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Step-by-step Solution

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1

Rewrite the expression $\frac{3x+5}{x^3-x^2-x+1}$ inside the integral in factored form

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

We can solve the integral $\int\frac{3x+5}{\left(x-1\right)^2\left(x+1\right)}dx$ 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

$u=x-1$
3

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

$du=dx$
4

Rewriting $x$ in terms of $u$

$x=u+1$
5

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

$\int\frac{8+3u}{u^2\left(2+u\right)}du$
6

Rewrite the fraction $\frac{8+3u}{u^2\left(2+u\right)}$ in $3$ simpler fractions using partial fraction decomposition

$\frac{8+3u}{u^2\left(2+u\right)}=\frac{A}{u^2}+\frac{B}{2+u}+\frac{C}{u}$
7

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 $u^2\left(2+u\right)$

$8+3u=u^2\left(2+u\right)\left(\frac{A}{u^2}+\frac{B}{2+u}+\frac{C}{u}\right)$
8

Multiplying polynomials

$8+3u=\frac{u^2\left(2+u\right)A}{u^2}+\frac{u^2\left(2+u\right)B}{2+u}+\frac{u^2\left(2+u\right)C}{u}$
9

Simplifying

$8+3u=\left(2+u\right)A+u^2B+u\left(2+u\right)C$
10

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

$\begin{matrix}8=2A&\:\:\:\:\:\:\:(u=0) \\ 2=4B&\:\:\:\:\:\:\:(u=-2) \\ 14=4A+4B+8C&\:\:\:\:\:\:\:(u=2)\end{matrix}$
11

Proceed to solve the system of linear equations

$\begin{matrix}2A & + & 0B & + & 0C & =8 \\ 0A & + & 4B & + & 0C & =2 \\ 4A & + & 4B & + & 8C & =14\end{matrix}$
12

Rewrite as a coefficient matrix

$\left(\begin{matrix}2 & 0 & 0 & 8 \\ 0 & 4 & 0 & 2 \\ 4 & 4 & 8 & 14\end{matrix}\right)$
13

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 4 \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & -\frac{1}{2}\end{matrix}\right)$
14

The integral of $\frac{8+3u}{u^2\left(2+u\right)}$ in decomposed fraction equals

$\int\left(\frac{4}{u^2}+\frac{1}{2\left(2+u\right)}+\frac{-1}{2u}\right)du$
15

Expand the integral $\int\left(\frac{4}{u^2}+\frac{1}{2\left(2+u\right)}+\frac{-1}{2u}\right)du$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{4}{u^2}du+\int\frac{1}{2\left(2+u\right)}du+\int\frac{-1}{2u}du$
16

The integral $\int\frac{4}{u^2}du$ results in: $\frac{-4}{x-1}$

$\frac{-4}{x-1}$
17

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

$\frac{1}{2}\ln\left(1+x\right)$
18

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

$-\frac{1}{2}\ln\left(x-1\right)$
19

Gather the results of all integrals

$\frac{-4}{x-1}+\frac{1}{2}\ln\left(1+x\right)-\frac{1}{2}\ln\left(x-1\right)$
20

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

$\frac{-4}{x-1}+\frac{1}{2}\ln\left(1+x\right)-\frac{1}{2}\ln\left(x-1\right)+C_0$

Final Answer

$\frac{-4}{x-1}+\frac{1}{2}\ln\left(1+x\right)-\frac{1}{2}\ln\left(x-1\right)+C_0$

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Function Plot

Plotting: $\frac{-4}{x-1}+\frac{1}{2}\ln\left(1+x\right)-\frac{1}{2}\ln\left(x-1\right)+C_0$

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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: 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.

Used Formulas

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