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I have the question "Using the substitution given, integrate with respect to $x$"

Here is my working:

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I understand everything expect for the bit after the word "hence".

If $6x^2$ cancels out, then where does the $\frac 16$ come from ?

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    Wait, didn't *you* write those equations? You write "Here is my working"? – 2017-02-02
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    Sorry I forgot to say I used the solutions to help with my working. – 2017-02-02

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The integral after the word hence is incorrect. It should read $$ \int x^2 U^5 \frac{1}{6x^2} du $$ which should make it clear where the 1/6 comes from. In fact, I'm not sure why there was a $6x^2$ instead of an $x^2$ in that integral in the first place.