How do you integrate by substitution $\int \frac{x^{2}}{{(16 - x^{3})}^{2}} d x$ $int x^2/(16-x^3)^2 dx$ ?

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Answer 1 · 2017-02-08T14:35:44

$\frac{1}{3 (16 - x^{3})} + C$ $1/(3(16 - x^3)) + C$

Let $u = 16 - x^{3}$ $u = 16 - x^3$ . Then $d u = - 3 x^{2} d x$ $du = -3x^2dx$ and $d x = \frac{d u}{- 3 x^{2}}$ $dx= (du)/(-3x^2)$ .

$\Rightarrow \int \frac{x^{2}}{u^{2}} \cdot \frac{d u}{- 3 x^{2}}$ $=>int x^2/u^2 * (du)/(-3x^2)$

$\Rightarrow - \frac{1}{3} \int \frac{1}{u^{2}} d u$ $=>-1/3int1/u^2 du$

$\Rightarrow - \frac{1}{3} \int u^{- 2} d u$ $=>-1/3int u^-2du$

$\Rightarrow - \frac{1}{3} (- u^{- 1}) + C$ $=>-1/3(-u^-1) + C$

$\Rightarrow \frac{1}{3 u} + C$ $=>1/(3u) + C$

$\Rightarrow \frac{1}{3 (16 - x^{3})} + C$ $=>1/(3(16 - x^3)) + C$

Hopefully this helps!

How do you integrate by substitution ∫x2(16−x3)2dxint x^2/(16-x^3)^2 dx?