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Integral of 6x to the third power plus 8 all to the 4th times 18x squared dx?
MollyHabby
This can easily be integrated by substitution.
Integrate[(6x3+8)418x2 dx]
let u=(6x3+8)
du/dx= 18x2
du=18x2 dx
Substituting u in gives...
Integrate [u4 du] = (1/5)(u5) + C
Substituting x back in gives..
(1/5)(6x3+8)5 + C
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∙ 14y ago
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