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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

We can factor the polynomial $x^3-x^2-x+1$ 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 $1$

$1$
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 $x^3-x^2-x+1$ will then be

$\pm1$
4

Trying all possible roots, we found that $1$ is a root of the polynomial. When we evaluate it in the polynomial, it gives us $0$ as a result

$1^3-1\cdot 1^2-1\cdot 1+1=0$
5

Now, divide the polynomial by the root we found $\left(x-1\right)$ 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 $1$. Add the result to the second coefficient and then multiply this by $1$ and so on

$\left|\begin{array}{c}1 & -1 & -1 & 1 \\ & 1 & 0 & -1 \\ 1 & 0 & -1 & 0\end{array}\right|1$
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 $\left(x-1\right)$

$\int\frac{3x+5}{\left(x^{2}-1\right)\left(x-1\right)}dx$
7

Factor the difference of squares $\left(x^{2}-1\right)$ as the product of two conjugated binomials

$\int\frac{3x+5}{\left(x+1\right)\left(x-1\right)\cdot \left(x-1\right)}dx$
8

When multiplying two powers that have the same base ($x-1$), you can add the exponents

$\int\frac{3x+5}{\left(x-1\right)^2\left(x+1\right)}dx$
9

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

$u=x-1$
10

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

$du=dx$
11

Rewriting $x$ in terms of $u$

$x=u+1$
12

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

$\int\frac{8+3u}{u^2\left(2+u\right)}du$
13

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

$\frac{8+3u}{u^2\left(2+u\right)}=\frac{A}{u^2}+\frac{B}{2+u}+\frac{C}{u}$
14

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

$8+3u=u^2\left(2+u\right)\left(\frac{A}{u^2}+\frac{B}{2+u}+\frac{C}{u}\right)$
15

Multiplying polynomials

$8+3u=\frac{u^2\left(2+u\right)A}{u^2}+\frac{u^2\left(2+u\right)B}{2+u}+\frac{u^2\left(2+u\right)C}{u}$
16

Simplifying

$8+3u=\left(2+u\right)A+u^2B+u\left(2+u\right)C$
17

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

$\begin{matrix}8=2A&\:\:\:\:\:\:\:(u=0) \\ 2=4B&\:\:\:\:\:\:\:(u=-2) \\ 14=4A+4B+8C&\:\:\:\:\:\:\:(u=2)\end{matrix}$
18

Proceed to solve the system of linear equations

$\begin{matrix}2A & + & 0B & + & 0C & =8 \\ 0A & + & 4B & + & 0C & =2 \\ 4A & + & 4B & + & 8C & =14\end{matrix}$
19

Rewrite as a coefficient matrix

$\left(\begin{matrix}2 & 0 & 0 & 8 \\ 0 & 4 & 0 & 2 \\ 4 & 4 & 8 & 14\end{matrix}\right)$
20

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 4 \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & -\frac{1}{2}\end{matrix}\right)$
21

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

$\int\left(\frac{4}{u^2}+\frac{1}{2\left(2+u\right)}+\frac{-1}{2u}\right)du$
22

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

$\int\frac{4}{u^2}du+\int\frac{1}{2\left(2+u\right)}du+\int\frac{-1}{2u}du$
23

We can solve the integral $\int\frac{1}{2\left(2+u\right)}du$ 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 $v$), which when substituted makes the integral easier. We see that $2+u$ it's a good candidate for substitution. Let's define a variable $v$ and assign it to the choosen part

$v=2+u$
24

Now, in order to rewrite $du$ in terms of $dv$, we need to find the derivative of $v$. We need to calculate $dv$, we can do that by deriving the equation above

$dv=du$
25

Substituting $v$ and $du$ in the integral and simplify

$\int\frac{4}{u^2}du+\frac{1}{2}\int\frac{1}{v}dv+\int\frac{-1}{2u}du$
26

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

$\frac{-4}{u}$
27

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

$\frac{1}{2}\ln\left(1+x\right)$
28

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

$-\frac{1}{2}\ln\left(u\right)$
29

Gather the results of all integrals

$\frac{-4}{u}+\frac{1}{2}\ln\left(1+x\right)-\frac{1}{2}\ln\left(u\right)$
30

Replace $u$ with the value that we assigned to it in the beginning: $x-1$

$\frac{-4}{x-1}+\frac{1}{2}\ln\left(1+x\right)-\frac{1}{2}\ln\left(u\right)$
31

Replace $u$ with the value that we assigned to it in the beginning: $x-1$

$\frac{-4}{x-1}+\frac{1}{2}\ln\left(1+x\right)-\frac{1}{2}\ln\left(x-1\right)$
32

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

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

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 ((3x+5)/(x^3+-1x^2))dx using partial fractionsSolve integral of ((3x+5)/(x^3+-1x^2))dx using basic integralsSolve integral of ((3x+5)/(x^3+-1x^2))dx using u-substitutionSolve integral of ((3x+5)/(x^3+-1x^2))dx using trigonometric substitution

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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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sin
cos
tan
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sec
csc

asin
acos
atan
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acsc

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cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
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Main Topic: Polynomial Factorization

They are a group of techniques that help us rewrite polynomial expressions as a product of factors.

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