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Integrate the function $\frac{\left(1-x^2\right)^2}{x^2+2x+1}$

Step-by-step Solution

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Final answer to the problem

$\frac{-\left(1-x\right)^{3}}{3}+C_0$
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Step-by-step Solution

Specify the solving method

1

Find the integral

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

We can solve the integral $\int\frac{\left(1-x^2\right)^2}{x^2+2x+1}dx$ by applying integration method of trigonometric substitution using the substitution

$x=\sin\left(\theta \right)$
3

Now, in order to rewrite $d\theta$ in terms of $dx$, we need to find the derivative of $x$. We need to calculate $dx$, we can do that by deriving the equation above

$dx=\cos\left(\theta \right)d\theta$
4

Substituting in the original integral, we get

$\int\frac{\left(1-\sin\left(\theta \right)^2\right)^2\cos\left(\theta \right)}{\sin\left(\theta \right)^2+2\sin\left(\theta \right)+1}d\theta$
5

Applying the trigonometric identity: $1-\sin\left(\theta \right)^2 = \cos\left(\theta \right)^2$

$\int\frac{\left(\cos\left(\theta \right)^2\right)^2\cos\left(\theta \right)}{\sin\left(\theta \right)^2+2\sin\left(\theta \right)+1}d\theta$
Why is 1 - sin(x)^2 = cos(x)^2 ?
6

Simplify $\left(\cos\left(\theta \right)^2\right)^2$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $2$

$\int\frac{\cos\left(\theta \right)^{4}\cos\left(\theta \right)}{\sin\left(\theta \right)^2+2\sin\left(\theta \right)+1}d\theta$
7

When multiplying exponents with same base you can add the exponents: $\cos\left(\theta \right)^{4}\cos\left(\theta \right)$

$\int\frac{\cos\left(\theta \right)^{5}}{\sin\left(\theta \right)^2+2\sin\left(\theta \right)+1}d\theta$
8

The trinomial $\sin\left(\theta \right)^2+2\sin\left(\theta \right)+1$ is a perfect square trinomial, because it's discriminant is equal to zero

$\Delta=b^2-4ac=2^2-4\left(1\right)\left(1\right) = 0$
9

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{\sin\left(\theta \right)^2}\:and\:b=\sqrt{1}$
10

Factoring the perfect square trinomial

$\int\frac{\cos\left(\theta \right)^{5}}{\left(\sin\left(\theta \right)+1\right)^{2}}d\theta$
11

Rewrite the trigonometric function $\cos\left(\theta \right)^{5}$ as the product of two lower exponents

$\int\frac{\cos\left(\theta \right)^{4}\cos\left(\theta \right)}{\left(\sin\left(\theta \right)+1\right)^{2}}d\theta$
12

Rewrite $\cos\left(\theta \right)^{4}$ as by applying trig identities

$\int\frac{\left(1-\sin\left(\theta \right)^2\right)^{2}\cos\left(\theta \right)}{\left(\sin\left(\theta \right)+1\right)^{2}}d\theta$
13

We can solve the integral $\int\frac{\left(1-\sin\left(\theta \right)^2\right)^{2}\cos\left(\theta \right)}{\left(\sin\left(\theta \right)+1\right)^{2}}d\theta$ 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 $\sin\left(\theta \right)$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=\sin\left(\theta \right)$
14

Now, in order to rewrite $d\theta$ 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

$du=\cos\left(\theta \right)d\theta$
15

Isolate $d\theta$ in the previous equation

$\frac{du}{\cos\left(\theta \right)}=d\theta$
16

Substituting $u$ and $d\theta$ in the integral and simplify

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

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

$\int\left(1-u\right)^{2}du$
18

We can solve the integral $\int\left(1-u\right)^{2}du$ 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 $1-u$ it's a good candidate for substitution. Let's define a variable $v$ and assign it to the choosen part

$v=1-u$
19

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

$dv=-1du$
20

Isolate $du$ in the previous equation

$dv=-du$
21

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

$\int-v^{2}dv$
22

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

$-\int v^{2}dv$
23

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2$

$\frac{-v^{3}}{3}$
24

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

$\frac{-\left(1-u\right)^{3}}{3}$
25

Replace $u$ with the value that we assigned to it in the beginning: $\sin\left(\theta \right)$

$\frac{-\left(1-\sin\left(\theta \right)\right)^{3}}{3}$
26

Express the variable $\theta$ in terms of the original variable $x$

$\frac{-\left(1-x\right)^{3}}{3}$
27

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

$\frac{-\left(1-x\right)^{3}}{3}+C_0$

Final answer to the problem

$\frac{-\left(1-x\right)^{3}}{3}+C_0$

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Function Plot

Plotting: $\frac{-\left(1-x\right)^{3}}{3}+C_0$

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9
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

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