How do you integrate $y = 3 {(5 x + 9)}^{9}$ $y=3 (5x+9)^9$ ?

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Answer 1 · 2016-11-12T22:35:27

Do a "u" substitution. Please see the explanation.

Let $u = 5 x + 9$ $u = 5x + 9$ , then $d u = 5 d x, d x = \frac{1}{5} d u$ $du = 5dx, dx = 1/5du$

$\int 3 {(5 x + 9)}^{9} d x = \frac{3}{5} \int u^{9} d u$ $int3(5x + 9)^9dx = 3/5intu^9du$

$\int 3 {(5 x + 9)}^{9} d x = \frac{3}{50} u^{10} + C$ $int3(5x + 9)^9dx = 3/50u^10 + C$

Reverse the substitution:

$\int 3 {(5 x + 9)}^{9} d x = \frac{3}{50} {(5 x + 9)}^{10} + C$ $int3(5x + 9)^9dx = 3/50(5x + 9)^10 + C$

How do you integrate y=3(5x+9)9y=3 (5x+9)^9?