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Core 3 Question; Did I get it right?

Integrate x sqrt (2x-1) dx

let u=2x-1

Got the answer as 1/4 [2/5 (2x-1) ^ 5/2 + 2/3 (2x-1) ^ 3/2 +C)

Taking the fractions outside the brackets..

1/15 [ (2x-1) ^5/2 + (2x-1) ^3/2 +C]

Is this correct?

Reply 1

Ab3rry

Nearly right - the answer itself is right but then the simplification with taking fractions out is wrong.

Try taking out 2 and a power of 2x-1.

Reply 2

MSI_10

Original post by Ab3rry

Nearly right - the answer itself is right but then the simplification with taking fractions out is wrong.

Try taking out 2 and a power of 2x-1.

Taking a 2 out from where, you mean from both fractions, 2/5 and 2/3?

Take a power of 2x-1 outside from the bracket to which it is raised to 5/2?

Reply 3

Ab3rry

Yeah, take the 2 out from both fractions.

Taking (2x-1)^5/2 out will give you a negative power in the brackets, which isn't ideal... so try the other option!

Reply 4

MSI_10

Original post by Ab3rry

Yeah, take the 2 out from both fractions.

Taking (2x-1)^5/2 out will give you a negative power in the brackets, which isn't ideal... so try the other option!

If the 2 is taken out from both fractions, it removes the 1/4 on the front of the eqn right?

so [ 1/5 (2x+1) ^5/2 + 1/3 (2x+1) ^3/2+C]

So remove a (2x+1) from (2x+) ^3/2?

that gives (2x+1) [ 1/5 (2x+1) ^ 5/2 + 1/3 (2x+1) ^1/2 +C] ?

EDIT: Replace all (2x+1) with (2x-1), whoops my bad :P