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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

$\frac{\left(1-x^2\right)^2}{-x-1}-\frac{4}{3}x^{3}-2x^2+C_0$
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Step-by-step Solution

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1

Find the integral

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

Rewrite the fraction $\frac{\left(1-x^2\right)^2}{x^2+2x+1}$ inside the integral as the product of two functions: $\left(1-x^2\right)^2\frac{1}{x^2+2x+1}$

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

We can solve the integral $\int\left(1-x^2\right)^2\frac{1}{x^2+2x+1}dx$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
4

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=\left(1-x^2\right)^2}\\ \displaystyle{du=-4\left(1-x^2\right)xdx}\end{matrix}$
5

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\frac{1}{x^2+2x+1}dx}\\ \displaystyle{\int dv=\int \frac{1}{x^2+2x+1}dx}\end{matrix}$
6

Solve the integral

$v=\int\frac{1}{x^2+2x+1}dx$
7

The trinomial $x^2+2x+1$ is a perfect square trinomial, because it's discriminant is equal to zero

$\Delta=b^2-4ac=2^2-4\left(1\right)\left(1\right) = 0$
8

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{x^2}\:and\:b=\sqrt{1}$
9

Factoring the perfect square trinomial

$\int\frac{1}{\left(x+1\right)^{2}}dx$
10

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

$u=x+1$
11

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

$du=dx$
12

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

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

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int u^{-2}du$
14

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 $-2$

$\frac{u^{-1}}{-1}$
15

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{\frac{1}{u}}{-1}$
16

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

$\frac{1}{-\left(x+1\right)}$
17

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

$\frac{\left(1-x^2\right)^2}{-x-1}+4\int\frac{\left(1-x^2\right)x}{-x-1}dx$
18

The integral $4\int\frac{\left(1-x^2\right)x}{-x-1}dx$ results in: $-2x^2-\frac{4}{3}x^{3}$

$-2x^2-\frac{4}{3}x^{3}$
19

Gather the results of all integrals

$\frac{\left(1-x^2\right)^2}{-x-1}-\frac{4}{3}x^{3}-2x^2$
20

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

$\frac{\left(1-x^2\right)^2}{-x-1}-\frac{4}{3}x^{3}-2x^2+C_0$

Final answer to the problem

$\frac{\left(1-x^2\right)^2}{-x-1}-\frac{4}{3}x^{3}-2x^2+C_0$

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Function Plot

Plotting: $\frac{\left(1-x^2\right)^2}{-x-1}-\frac{4}{3}x^{3}-2x^2+C_0$

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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

How to improve your answer:

Main Topic: Integral Calculus

Integration assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data.

Used Formulas

See formulas (11)

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