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C3 - Differentiation

[SOLVED]

I don't know how to begin solving the second part of the following question:

i. Given that y=(x+1)2x1y = (x + 1)\sqrt{2x - 1} show that dydx\frac{dy}{dx} can be written in the form:

kx2x1\displaystyle \frac{kx}{\sqrt{2x - 1}} and state the value of k.

(This part is fine and I got k = 3)

ii. Hence evaluate

15x2x1 dx\displaystyle \int_1^5\frac{x}{\sqrt{2x - 1}}\ dx

I do not know how to begin this part. This question is from a differentiation section of the textbook I'm using and so far I have learned only the chain and product rules and do not know how to integrate this so I assume that I am expected to solve this from the first part of the question. If someone could point me in the right direction I would appreciate it.
(edited 11 years ago)
Remember that integration reverses differentiation.
Reply 2
Original post by Classical Liberal
Remember that integration reverses differentiation.


15x2x1 dx\displaystyle \int_1^5\frac{x}{\sqrt{2x - 1}}\ dx

=13((5+1)2(5)1(1+1)2(1)1)=163= \displaystyle \frac{1}{3}((5 + 1)\sqrt{2(5) - 1} - (1 + 1)\sqrt{2(1) - 1}) = \frac{16}{3}

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