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`int (x^3+x+1)/(x^4+2x^2+1) dx` Find the indefinite integral

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` 

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