integrate (x 9 +x 6 +x 3 )(2x 6 +3x 3 +6) 1/3 ans= (1/24 )( 2x 9 +3x 6 +6x 3 ) 4/3 + c

Question

Vikas TU · Answer

Dear student
I = integral of (x^8 + x^5 + x^2) (2x^9+ 3x^6 + 6x^3)^1/3 dx
Let 2x^9+ 3x^6 + 6x^3 = t
2(2x^8+ 3x^5 + 6x^2) = dt
Put this in the value of I

ans= (1/24 )(2x⁹+3x⁶+6x³)^4/3 + c

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integrate (x 9 +x 6 +x 3 )(2x 6 +3x 3 +6) 1/3 ans= (1/24 )( 2x 9 +3x 6 +6x 3 ) 4/3 + c

integrate (x⁹+x⁶+x³)(2x⁶+3x³+6)^1/3

ans= (1/24 )(2x⁹+3x⁶+6x³)^4/3 + c

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integrate (x 9 +x 6 +x 3 )(2x 6 +3x 3 +6) 1/3 ans= (1/24 )( 2x 9 +3x 6 +6x 3 ) 4/3 + c

integrate (x9+x6+x3)(2x6+3x3+6)1/3 ans= (1/24 )(2x9+3x6+6x3)4/3 + c

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integrate (x⁹+x⁶+x³)(2x⁶+3x³+6)^1/3

ans= (1/24 )(2x⁹+3x⁶+6x³)^4/3 + c