How do you integrate by substitution $\int x^{3} {(x^{4} + 3)}^{2} d x$ $int x^3(x^4+3)^2 dx$ ?

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Answer 1 · 2016-10-24T06:07:31

Please see the explanation.

Let $u = x^{4} + 3$ $u = x^4 + 3$ , then $d u = 4 x^{3} d x$ $du = 4x^3dx$ or $x^{3} d x = (\frac{1}{4}) d u$ $x^3dx = (1/4)du$

$\int x^{3} {(x^{4} + 3)}^{2} d x = \frac{1}{4} \int u^{2} d u$ $intx^3(x^4+3)^2dx = 1/4intu^2du$

Integrate using the power rule:

$\int x^{3} {(x^{4} + 3)}^{2} d x = (\frac{1}{12}) u^{3} + C$ $intx^3(x^4+3)^2dx = (1/12)u^3 + C$

Reverse the substitution:

$\int x^{3} {(x^{4} + 3)}^{2} d x = (\frac{1}{12}) {(x^{4} + 3)}^{3} + C$ $intx^3(x^4+3)^2dx = (1/12)(x^4 + 3)^3 + C$

How do you integrate by substitution int x^3(x^4+3)^2 dx∫x3(x4+3)2dxint x^3(x^4+3)^2 dx?