How solve it ? PLEASE use Integration by Change of Variable U Substitution

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How solve it ? PLEASE use Integration by Change of Variable U Substitution

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Answer 1 · 2017-06-07T01:30:44

Let $u = x^{3} - 3 x + 16$ $u = x^3-3x+16$ (the argument of the radical).

Explanation:

Then $d u = (3 x^{2} - 3) d x = 3 (x^{2} - 1) d x$ $du = (3x^2-3)dx = 3(x^2-1)dx$

$\int \frac{x^{2} - 1}{\sqrt[3]{x^{3} - 3 x + 16}} d x = \frac{1}{3} \int {(x^{3} - 3 x + 16)}^{- \frac{1}{3}} \cdot 3 (x^{2} - 1) d x$ $int (x^2-1)/root(3)(x^3-3x+16) dx = 1/3 int (x^3-3x+16)^(-1/3) * 3(x^2-1)dx$

$= \frac{1}{3} \int u^{- \frac{1}{3}} d u = \frac{1}{3} (\frac{3}{2} u^{\frac{2}{3}}) + C$ $= 1/3 int u^(-1/3) du = 1/3(3/2 u^(2/3))+C$

$= \frac{1}{2} \sqrt[3]{{(x^{3} - 3 x + 16)}^{2}} + C$ $=1/2 root(3)((x^3-3x+16)^2) + C$

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How solve it ? PLEASE use Integration by Change of Variable U Substitution

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