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Question #9ae03

1 Answer

Feb 5, 2017

# 5/8(root5(y^4-8))^8+C.#.

Explanation:

We will use the Substitution #: root5(y^4-8)=x#.

#rArr (y^4-8)=x^5 rArr d(y^4-8)=d(x^5), i.e., 4y^3dy=5x^4dx#.

#"Therefore, "I=int(root5(y^4-8))^3(4y^3dy)#

#=int(x^3)(5x^4)dx=5intx^7dx=5(x^(7+1)/(7+1))=5/8x^8#.

#rArr I=5/8(root5(y^4-8))^8+C.#.

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