Integral [x^3 + 1]^2 dx = ? I found that it was [(x^3 + 1)^3] / 9x^2, but that is wrong? Details inside.

Integral [x^3 + 1]^2 dx = ? I found that it was [(x^3 + 1)^3] / 9x^2, but that is wrong? Details inside.
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1 Answer

Answer :

Expand this using FOIL. You will get three distinct terms and it will be a very easy integral. The answer should be (x^7)/7 + (x^4)/2 + x
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