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

Step-by-step Solution

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Final answer to the problem

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

How should I solve this problem?

  • Integrate by parts
  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
  • FOIL Method
  • Load more...
Can't find a method? Tell us so we can add it.
1

Rewrite the expression $\frac{x+3}{x^2+4x+5}$ inside the integral in factored form

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

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

$u=x+2$
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 $u$

$x=u-2$
5

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

$\int\frac{u+1}{u^2+1}du$
6

Expand the fraction $\frac{u+1}{u^2+1}$ into $2$ simpler fractions with common denominator $u^2+1$

$\int\left(\frac{u}{u^2+1}+\frac{1}{u^2+1}\right)du$
7

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

$\int\frac{u}{u^2+1}du+\int\frac{1}{u^2+1}du$
8

Rewrite the fraction $\frac{u}{u^2+1}$ inside the integral as the product of two functions: $u\frac{1}{u^2+1}$

$\int u\frac{1}{u^2+1}du+\int\frac{1}{u^2+1}du$
9

We can solve the integral $\int u\frac{1}{u^2+1}du$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
10

First, identify or choose $u$ and calculate it's derivative, $du$

$\begin{matrix}\displaystyle{u=u}\\ \displaystyle{du=du}\end{matrix}$
11

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\frac{1}{u^2+1}du}\\ \displaystyle{\int dv=\int \frac{1}{u^2+1}du}\end{matrix}$
12

Solve the integral to find $v$

$v=\int\frac{1}{u^2+1}du$
13

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

$\frac{1}{\sqrt{1}}\arctan\left(\frac{u}{\sqrt{1}}\right)$
14

Now replace the values of $u$, $du$ and $v$ in the last formula

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

Replace $u$ with the value that we assigned to it in the beginning: $x+2$

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

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

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

Gather the results of all integrals

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

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

$\arctan\left(x+2\right)$
19

Gather the results of all integrals

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

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

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

Final answer to the problem

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

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Function Plot

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

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a
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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

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