**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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