- #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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