Wednesday, December 29, 2021

12 Days of Christmas Integrals: ∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

12 Days of Christmas Integrals:  ∫ 2 ∙ x ∙ (x^2 + 1)^2 dx


On the Fifth day of Christmas Integrals, the integral featured today is...


∫ 2 ∙ x ∙ (x^2 + 1)^2 dx


I am going to use two approaches and show why both approaches are valid.  Remember we are working indefinite integrals.  


Substitution Method


∫ 2 ∙ x ∙ (x^2 + 1)^2 dx


u = x^2 + 1 

du = 2 ∙ x dx


u^2 = (x^2 + 1)^2


∫ u^2 du


= u^3 ÷ 3 + C


= (x^2 + 1)^3 ÷ 3 + C


= (x^6 + 3 ∙ x^4 + 3 ∙ x^2 + 1) ÷ 3 + C


= (x^6 + 3 ∙ x^4 + 3 ∙ x^2) ÷ 3 + 1 ÷ 3 + C


= x^6 ÷ 3 +  x^4 +  x^2 + 1 ÷ 3 + C



Algebraic Method


∫ 2 ∙ x ∙ (x^2 + 1)^2 dx


= ∫ 2 ∙ x ∙ (x^4 + 2 ∙ x^2 + 1) dx


= ∫ 2 ∙ x^5 + 4 ∙ x^3 + 2 ∙ x dx


= 2/6 ∙ x^6 + 4/4 ∙ x^4 + 2/2 ∙ x^2 + C


= x^6 ÷ 3 + x^4 + x^2 + C



Why are the two methods acceptable?


In indefinite integration, a "+ C" is added to show that the solutions to indefinite integrals are a class of functions.   C is an arbitrary numerical constant.


d/dx[  f(x) + C ] =   d/dx [ f(x) ] + 0  = d/dx [ f(x) ]


Since C is a numerical constant,  C + 1/3 is also a numerical constant.  




Eddie 


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