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what is the indefinite integral of 1/x^3
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1/x^3= x^-3.....integral x^n is (x^(n+1))/(n+1)
just to make pretty what hartnn wrote\[\large\frac1{x^3}=x^{-3}\]and\[\large\int x^ndx=\frac{x^{n+1}}{n+1}+C\]
thanks ...i've got it
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what about the indefinite integral of y^3(2y^2-3)
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distribute, then apply the same rule but please post each Q separately thanks
multiply y^3 inside.....which gives 2y^5-3y^3...then apply same formula to both terms
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