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

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Answer 1 · 2016-12-27T07:47:12

The answer is $= \frac{{(3 x - 9)}^{10}}{30} + C$ $=(3x-9)^10/30+C$

We need

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx=x^(n+1)/(n+1)+C$

We do this integral by substitution

Let $u = 3 x - 9$ $u=3x-9$ , $\Rightarrow$ $=>$ , $d u = 3 d x$ $du=3dx$

Therefore,

$\int {(3 x - 2)}^{9} d x = \frac{1}{3} \int u^{9} d u$ $int(3x-2)^9dx=1/3intu^9du$

$= \frac{1}{3} \frac{u^{10}}{10}$ $=1/3u^10/10$

$= \frac{{(3 x - 9)}^{10}}{30} + C$ $=(3x-9)^10/30+C$

How do you integrate ∫(3x−2)9int (3x-2)^9 using substitution?