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junior college 1 | H1 Maths
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Observe that differentiating x^3 gives us 3x^2, which is a constant multiple of 6x^2.
If you have done differentiation before, if we were to differentiate e^(x^3), we would get e^(x^3) times d/dx (x^3)...
...where the d/dx (x^3) = 3x^2 is our chain.
So, done in reverse, integrating 3x^2 * e^(x^3) is e^(x^3) + c.
Therefore, integrating the integral in part a, we get 2e^(x^3) + c.
This is why I made mention of the "reverse-chain".
Substitution is more difficult, and I can think of what substitutions to use.
Perhaps the first one can use some ln expression (like x^3) = ln u, but I am not 100% sure.
Second one maybe is using u = x^3 - 3.
Third one, the toughest, requires me to convert 2x^2 to 4 tan^2 x so that I can simplify the denominator using the identity tan^2 x + 1 = sec^2 x.
Having said that, the reverse chain idea is more straightforward as all three questions are capable of being done this way.
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Part a