Evaluate Int [(x^4 -1) ^2*4x^3 dx].

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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Int (x^4-1)^2 * 4x^3 dx

Let u = x^4  ==> du = 4x^3 dx

==> intg (x^4 -1)^2 ( 4x^3) dx = intg (u-1)^2 * du.

Let us integrate with respect to u.

==> intg (u-1)^2 du = intg (u^2 - 2u + 1) du

                               = intg u^2 du - intg 2u du + intg 1 du.

                                = u^3/3 - 2u^2/2  + u + C

==> intg (u-1)^2 du = (1/3)u^3 - u^2 + u + c

Now we will substitute with u= x^4

==> intg (u^2-1)du = (1/3)(x^4)^3 - (x^4)^2 + x^4 + C

                              = (1/3)x^12 - x^8 + x^4 + C

==> intg (x^4-1)^2 * 3x^2 dx = (1/3)x^12 - x^8 + x^4 + C

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to evaluate Int [(x^4 -1) ^2*4x^3 dx]

Let us first write x^4 – 1 = y

For y = x^4 – 1, we have dy/dx = 4x^3

=> dy = 4x^3 dx

Now we can rewrite Int [(x^4 -1) ^2*4x^3 dx]

=> Int [y^2 dy]

We know that Int [x^n dx] = [x^ (n+1)]/ (n +1)

=> y^3 / 3 +C

Now replace y with x^4 – 1 again.

=> (x^4 – 1) ^3 /3 +C

Therefore Int [(x^4 -1) ^2*4x^3 dx] = (x^4 – 1) ^3 /3 + C

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