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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