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# x^8(10+x^9)^3 Calculate the indefinite integral?

### 4 Answers

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- 2 months agoFavorite Answer
x^8 * (10 + x^9)^3 * dx

u = 10 + x^9

du = 9x^8 * dx

(1/9) * 9 * x^8 * dx * (10 + x^9)^3 =>

(1/9) * du * u^3 =>

(1/9) * u^3 * du

Integrate

(1/9) * (1/4) * u^4 + C =>

(1/36) * (10 + x^9)^4 + C

- PhilipLv 62 months ago
J = Int'l of x^8(10+x^9)^3dx.;

Put u = 10+x^9. Then du/dx = 9x^8, ie., (1/9)du=(x^8)dx & J=(1/9)Int'l(u^3du)

= (1/36)u^4 + c, where c is an arbitrary constant of integration. Then indefinite

integral is (1/36)(10+x^9)^4 + c.

- Ian HLv 72 months ago
Guess and correct like this

Let y = (x^9 + 10)^4

dy/dx = 36x^8(x^9 + 10)^3

I = ∫x^8(10 + x^9)^3 dx = (1/36)(10 + x^9)^4 + C

- micatkieLv 72 months ago
Let u = 10 + x⁹

Then, du = 9x⁸dx, and thus x⁸dx = (1/9)du

∫ x⁸ (10 + x⁹)³ dx

= ∫ (10 + x⁹)³ (x⁸dx)

= (1/9) ∫ u³ du

= (1/9) (1/4) u⁴ + C

= (1/36)(10 + x⁹)⁴ + C

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