# Integral of (x+3)/(4x+5+x^2)

## \int\frac{x+3}{x^2+4x+5}dx

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$\frac{1}{2}\ln\left|4x+x^2+5\right|+arctan\left(2+x\right)+C_0$

## Step by step solution

Problem

$\int\frac{x+3}{x^2+4x+5}dx$
1

Split the fraction $\frac{x+3}{5+4x+x^2}$ in two terms with same denominator

$\int\left(\frac{3}{5+4x+x^2}+\frac{x}{5+4x+x^2}\right)dx$
2

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

$\int\frac{3}{5+4x+x^2}dx+\int\frac{x}{5+4x+x^2}dx$
3

Add and subtract $\displaystyle\left(\frac{b}{2a}\right)^2$

$\int\frac{3}{-4+4+5+4x+x^2}dx+\int\frac{x}{5+4x+x^2}dx$
4

Factor the perfect square trinomial $x^2+4x+4$

$\int\frac{3}{-4+5+\left(2+x\right)^2}dx+\int\frac{x}{5+4x+x^2}dx$
5

Subtract the values $5$ and $-4$

$\int\frac{3}{\left(2+x\right)^2+1}dx+\int\frac{x}{5+4x+x^2}dx$
6

Solve the integral $\int\frac{3}{\left(2+x\right)^2+1}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=2+x \\ du=dx\end{matrix}$
7

Substituting $u$ and $dx$ in the integral

$\int\frac{3}{u^2+1}du+\int\frac{x}{5+4x+x^2}dx$
8

Taking the constant out of the integral

$3\int\frac{1}{u^2+1}du+\int\frac{x}{5+4x+x^2}dx$
9

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

$3\cdot 1arctan\left(u\right)+\int\frac{x}{5+4x+x^2}dx$
10

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

$3arctan\left(2+x\right)+\int\frac{x}{5+4x+x^2}dx$
11

Add and subtract $\displaystyle\left(\frac{b}{2a}\right)^2$

$3arctan\left(2+x\right)+\int\frac{x}{-4+4+5+4x+x^2}dx$
12

Factor the perfect square trinomial $x^2+4x+4$

$3arctan\left(2+x\right)+\int\frac{x}{-4+5+\left(2+x\right)^2}dx$
13

Subtract the values $5$ and $-4$

$3arctan\left(2+x\right)+\int\frac{x}{\left(2+x\right)^2+1}dx$
14

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

$\begin{matrix}u=2+x \\ du=dx\end{matrix}$
15

Rewriting $x$ in terms of $u$

$x=u-2$
16

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

$3arctan\left(2+x\right)+\int\frac{u-2}{u^2+1}du$
17

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

$3arctan\left(2+x\right)+\int\left(\frac{-2}{u^2+1}+\frac{u}{u^2+1}\right)du$
18

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

$3arctan\left(2+x\right)+\int\frac{-2}{u^2+1}du+\int\frac{u}{u^2+1}du$
19

Taking the constant out of the integral

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

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

$3arctan\left(2+x\right)-2\cdot 1arctan\left(u\right)+\int\frac{u}{u^2+1}du$
21

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

$3arctan\left(2+x\right)-2arctan\left(2+x\right)+\int\frac{u}{u^2+1}du$
22

Adding $-2arctan\left(2+x\right)$ and $3arctan\left(2+x\right)$

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

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

$\begin{matrix}v=u^2+1 \\ dv=2udu\end{matrix}$
24

Isolate $du$ in the previous equation

$\frac{dv}{2u}=du$
25

Substituting $v$ and $du$ in the integral

$\int\frac{1}{2v}dv+arctan\left(2+x\right)$
26

Taking the constant out of the integral

$\frac{1}{2}\int\frac{1}{v}dv+arctan\left(2+x\right)$
27

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

$\frac{1}{2}\cdot 1\ln\left|v\right|+arctan\left(2+x\right)$
28

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

$\frac{1}{2}\ln\left|u^2+1\right|+arctan\left(2+x\right)$
29

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

$\frac{1}{2}\ln\left|\left(2+x\right)^2+1\right|+arctan\left(2+x\right)$
30

Expanding the polynomial

$\frac{1}{2}\ln\left|4x+x^2+5\right|+arctan\left(2+x\right)$
31

$\frac{1}{2}\ln\left|4x+x^2+5\right|+arctan\left(2+x\right)+C_0$

$\frac{1}{2}\ln\left|4x+x^2+5\right|+arctan\left(2+x\right)+C_0$

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