# Integral of (x^2+3)/((x+1)(x^2+1))

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

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

## Step by step solution

Problem

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

Using partial fraction decomposition, the fraction $\frac{3+x^2}{\left(1+x^2\right)\left(1+x\right)}$ can be rewritten as

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

Now we need to find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $\left(1+x^2\right)\left(1+x\right)$

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

Multiplying polynomials

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

Simplifying

$3+x^2=\left(1+x\right)\left(A+Bx\right)+C\left(1+x^2\right)$
5

Expand the polynomial

$3+x^2=C+Cx^2+A+Bx+Ax+Bx^2$
6

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

$\begin{matrix}4=2C&\:\:\:\:\:\:\:(x=-1) \\ 4=2C+2A+2B&\:\:\:\:\:\:\:(x=1) \\ 7=A+2B+2A+4B+5C&\:\:\:\:\:\:\:(x=2)\end{matrix}$
7

Proceed to solve the system of linear equations

$\begin{matrix}0A & + & 0B & + & 2C & =4 \\ 2A & + & 2B & + & 2C & =4 \\ 3A & + & 6B & + & 5C & =7\end{matrix}$
8

Rewrite as a coefficient matrix

$\left(\begin{matrix}0 & 0 & 2 & 4 \\ 2 & 2 & 2 & 4 \\ 3 & 6 & 5 & 7\end{matrix}\right)$
9

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2\end{matrix}\right)$
10

The decomposed integral equivalent is

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

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int\frac{1-x}{1+x^2}dx+\int\frac{2}{1+x}dx$
12

Apply the formula: $\int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|$, where $b=1$ and $n=2$

$\int\frac{1-x}{1+x^2}dx+2\ln\left|1+x\right|$
13

Split the fraction $\frac{-x+1}{1+x^2}$ in two terms with same denominator

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

The integral of a sum of two or more functions is equal to the sum of their integrals

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

Taking the constant out of the integral

$2\ln\left|1+x\right|+\int\frac{1}{1+x^2}dx-\int\frac{x}{1+x^2}dx$
16

Solve the integral applying the formula $\displaystyle\int\frac{x'}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)$

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

Solve the integral $\int\frac{x}{1+x^2}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=1+x^2 \\ du=2xdx\end{matrix}$
18

Isolate $dx$ in the previous equation

$\frac{du}{2x}=dx$
19

Substituting $u$ and $dx$ in the integral

$2\ln\left|1+x\right|+arctan\left(x\right)-\int\frac{1}{2u}du$
20

Taking the constant out of the integral

$2\ln\left|1+x\right|+arctan\left(x\right)-1\cdot \frac{1}{2}\int\frac{1}{u}du$
21

Multiply $\frac{1}{2}$ times $-1$

$2\ln\left|1+x\right|+arctan\left(x\right)-\frac{1}{2}\int\frac{1}{u}du$
22

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$2\ln\left|1+x\right|+arctan\left(x\right)-\frac{1}{2}\cdot 1\ln\left|u\right|$
23

Substitute $u$ back for it's value, $1+x^2$

$2\ln\left|1+x\right|+arctan\left(x\right)-\frac{1}{2}\ln\left|1+x^2\right|$
24

Add the constant of integration

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

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

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### Main topic:

Integrals by partial fraction expansion

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