How do you integrate by substitution $\int {(x^{2} - 9)}^{3} (2 x) d x$ $int(x^2-9)^3(2x)dx$ ?

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How do you integrate by substitution $\int {(x^{2} - 9)}^{3} (2 x) d x$ $int(x^2-9)^3(2x)dx$ ?

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Answer 1 · 2016-11-10T14:53:15

$\frac{{(x^{2} - 9)}^{4}}{4} + c$ $(x^2-9)^4/4+c$

$y = x^{2} - 9$ $y=x^2-9$

Explanation:

$y = x^{2} - 9$ $y=x^2-9$

$d y = 2 x d x$ $dy=2xdx$

$\int {(x^{2} - 9)}^{3} (2 x) d x = \int y^{3} d y = \frac{y^{4}}{4} + c = \frac{{(x^{2} - 9)}^{4}}{4} + c$ $int(x^2-9)^3(2x)dx=inty^3dy=y^4/4+c=(x^2-9)^4/4+c$

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How do you integrate by substitution $\int {(x^{2} - 9)}^{3} (2 x) d x$ $int(x^2-9)^3(2x)dx$ ?

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Explanation:

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How do you integrate by substitution ∫(x2−9)3(2x)dxint(x^2-9)^3(2x)dx?

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How do you integrate by substitution $\int {(x^{2} - 9)}^{3} (2 x) d x$ $int(x^2-9)^3(2x)dx$ ?