How do you integrate $\int 12 x^{3} {(3 x^{4} + 4)}^{4}$ $int12x^3(3x^4+4)^4$ using substitution?

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Answer 1 · 2016-10-30T19:25:01

Please see the explanation

Given

$\int 12 x^{3} {(3 x^{4} + 4)}^{4} d x$ $int12x^3(3x^4 + 4)^4dx$

let $u = 3 x^{4} + 4$ $u = 3x^4 + 4$ , then $d u = 12 x^{3} d x$ $du = 12x^3dx$

$\int 12 x^{3} {(3 x^{4} + 4)}^{4} d x = \int u^{4} d u = (\frac{1}{5}) u^{5} + C$ $int12x^3(3x^4 + 4)^4dx = intu^4du = (1/5)u^5 + C$

Reverse the substitution:

$\int 12 x^{3} {(3 x^{4} + 4)}^{4} d x = (\frac{1}{5}) {(3 x^{4} + 4)}^{5} + C$ $int12x^3(3x^4 + 4)^4dx = (1/5)(3x^4 + 4)^5 + C$

How do you integrate int12x^3(3x^4+4)^4∫12x3(3x4+4)4int12x^3(3x^4+4)^4 using substitution?