Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 34

Indefinite integral are written in the form of int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function of f(x)
C as the arbitrary constant known as constant of integration

To determine the indefinite integral of int (x^3+x+1)/(x^4+2x^2+1) dx , we apply partial fraction decomposition to expand the integrand: f(x)=(x^3+x+1)/(x^4+2x^2+1) .
The pattern on setting up partial fractions will depend on the factors of the denominator. For the given problem, the denominator is in a similar form of perfect squares trinomial: x^2+2xy+y^2= (x+y)^2
Applying the special factoring on (x^4+2x^2+1) , we get: (x^4+2x+1)= (x^2+1)^2 .
For the repeated quadratic factor (x^2+1)^2 , we will have partial fraction: (Ax+B)/(x^2+1) +(Cx+D)/(x^2+1)^2 .
The integrand becomes:
(x^3+x+1)/(x^4+2x^2+1)=(Ax+B)/(x^2+1) +(Cx+D)/(x^2+1)^2
Multiply both sides by the LCD =(x^2+1)^2 :
((x^3+x+1)/(x^4+2x^2+1)) *(x^2+1)^2=((Ax+B)/(x^2+1) +(Cx+D)/(x^2+1)^2)*(x^2+1)^2
x^3+x+1=(Ax+B)(x^2+1) +Cx+D
x^3+x+1=Ax^3 +Ax+Bx^2+B+Cx+D
x^3+0x^2 + x+1=Ax^3 +Ax+Bx^2+B+Cx+D
Equate the coefficients of similar terms on both sides to list a system of equations:
Terms with x^3 : 1 = A
Terms with x^2 : 0=B
Terms with x : 1 = A+C
Plug-in A =1 on 1 =A+C , we get:
1 =1+C
C =1-1
C =0
Constant terms: 1=B+D
Plug-in B =0 on 1 =B+D , we get:
1 =0+D
D =1
Plug-in the values of A =1 , B=0 , C=0 , and D=1 , we get the partial fraction decomposition:
(x^3+x+1)/(x^4+2x^2+1)=(1x+0)/(x^2+1) +(0x+1)/(x^2+1)^2
=x/(x^2+1) +1/(x^2+1)^2
Then the integral becomes:
int (x^3+x+1)/(x^4+2x^2+1) dx = int [x/(x^2+1) +1/(x^2+1)^2] dx
Apply the basic integration property: int (u+v) dx = int (u) dx +int (v) dx.
int [x/(x^2+1) +1/(x^2+1)^2] dx=int x/(x^2+1)dx +int 1/(x^2+1)^2 dx
For the first integral, we apply integration formula for rational function as:
int u /(u^2+a^2) du = 1/2ln|u^2+a^2|+C
Then, int x/(x^2+1)dx=1/2ln|x^2+1|+C or (ln|x^2+1|)/2+C
For the second integral, we apply integration by trigonometric substitution.
We let x = tan(u) then dx= sec^2(u) du
Plug-in the values, we get:
int 1/(x^2+1)^2 dx = int 1 /(tan^2(u)+1)^2 * sec^2(u) du
Apply the trigonometric identity: tan^2(u) +1 = sec^2(u) and trigonometric property: 1/(sec^2(u)) =cos^2(u)
int 1 /(tan^2(u)+1)^2 * sec^(u) du =int 1 /(sec^2(u))^2 * sec^2(u) du
= int 1 /(sec^4(u)) * sec^2(u) du
=int 1/(sec^2(u)) du
= int cos^2(u) du
Apply the integration formula for cosine function: int cos(x) dx = 1/2[x+sin(x)cos(x)]+C
int cos^2(u) du= 1/2[u+sin(u)cos(u)]+C
Based from x= tan(u) then :
u =arctan(x)
sin(u) = x/sqrt(x^2+1)
cos(u) =1/sqrt(x^2+1)
Then the integral becomes:
int 1/(x^2+1)^2dx
= 1/2[arctan(x) + (x/sqrt(x^2+1))*(1/sqrt(x^2+1))]
=arctan(x)/2+x/(2x^2+2)
Combining the results, we get:
int (x^3+x+1)/(x^4+2x^2+1) dx =(ln|x^2+1|)/2+arctan(x)/2+x/(2x^2+2)+C