<s>[INST] Possible division by zero in integration by u-substitution When I was solving the indefinite integral $\int x^2(x^3-7)^3\ dx$, my working was as follows:
Let $u=x^3-7$, then $dx=\frac{1}{3x^2}du$.
Hence, $\int x^2u^3\frac{1}{3x^2}\ du=\frac{1}{3}\int u^3\ du=\frac{u^4}{12}+C=\frac{(x^3-7)^4}{12}+C$.
However, during the u-substitution we get $dx=\frac{1}{3x^2}du$. If $x=0$, then wouldn't that mean we would be diving by zero?
So when we go ahead to find a definite integral with $x=0$ in the limit, why is it okay to ignore the fact that we are dividing by zero?
 [/INST] You may avoid dividing by zero if instead of $$dx=\frac{1}{3x^2}du$$ consider $$ du = 3x^2 dx $$ and substitute $$ x^2dx = \frac {1}{3} du$$
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