- #1

DH

- 21

- 0

Determine the integral:

[tex]y = \int_{0}^{1} 1/(u^4+1)du [/tex]

and

[tex]y = \int_{0}^{1} 1/(u^5+1)du [/tex]

and

[tex]y = \int_{0}^{1} 1/(u^6+1)du [/tex]

[tex]y = \int_{0}^{1} 1/(u^4+1)du [/tex]

and

[tex]y = \int_{0}^{1} 1/(u^5+1)du [/tex]

and

[tex]y = \int_{0}^{1} 1/(u^6+1)du [/tex]

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