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Integral of a Algebraic Fraction Help

Hi so I'm working through past papers and came across this...

Unparseable latex formula:

[br]\int-\dfrac{3}{2x},dx\[br]



I thought the answer was...

[br]−32ln⁡∣2x∣+c[br][br]-\dfrac{3}{2} \ln |2x| +c[br]

But the mark scheme has...

[br]−32ln⁡∣x∣+c[br][br]-\dfrac{3}{2} \ln |x| +c[br]

Can someone explain this please? It's for CCEA C3.

Thanks :smile:
(edited 10 years ago)
Reply 1
Avatar for joostan
joostan
13
Original post by 01Chris02
Hi so I'm working through past papers and came across this...

Unparseable latex formula:

[br]\int-\dfrac{3}{2x},dx\[br]



I thought the answer was...

[br]−32ln⁡∣2x∣+c[br][br]-\dfrac{3}{2} \ln |2x| +c[br]

But the mark scheme has...

[br]−32ln⁡∣2∣+c[br][br]-\dfrac{3}{2} \ln |2| +c[br]

Can someone explain this please? It's for CCEA C3.

Thanks :smile:

The mark scheme is wrong. :smile:
Reply 2
Avatar for qgujxj39
qgujxj39
17
You're right, looks like a typo in the mark scheme.
Reply 3
Avatar for hoodboilu4
hoodboilu4
9
I actual fact both answers are wrong.

Remember that you can take multiples outside of integrals:

∫−32xdx=−32∫1xdx\int -\dfrac{3}{2x} dx = -\dfrac{3}{2} \int \dfrac{1}{x} dx

Then you can intergrate 1x\dfrac{1}{x} and multiply by −32- \dfrac{3}{2}

−32∫1xdx= -\dfrac{3}{2}\int \dfrac{1}{x} dx =



−32∗ln(x)+C= - \dfrac{3}{2} * ln(x) + C =



−32ln(x)+C-\dfrac{3}{2}ln(x)+C
Reply 4
Avatar for 01Chris02
01Chris02
OP
13
Is this one of those situations where are the above are right but the constant changes to account for the difference?
Thanks by the way!
Reply 5
Avatar for joostan
joostan
13
Original post by hoodboilu4
.

Original post by 01Chris02
Is this one of those situations where are the above are right but the constant changes to account for the difference?
Thanks by the way!


They're not both wrong.
recall that: −32ln⁡(2x)=−32ln⁡(2)+−32ln⁡(x)-\frac{3}{2}\ln(2x) = -\frac{3}{2}\ln(2) + -\frac{3}{2}\ln(x)
And also that:
−32ln⁡(2)-\frac{3}{2}\ln(2)
is just a constant.
You simply get a different constant (+c) after you integrate. :smile:
(edited 10 years ago)
Reply 6
Avatar for 01Chris02
01Chris02
OP
13
Oh wait sorry I just realised that I typed that in wrong the mark scheme says...
[br]−32ln⁡(x)+c[br][br]-\dfrac{3}{2} \ln (x) +c[br]
Reply 7
Avatar for joostan
joostan
13
Original post by 01Chris02
Oh wait sorry I just realised that I typed that in wrong the mark scheme says...
[br]−32ln⁡(x)+c[br][br]-\dfrac{3}{2} \ln (x) +c[br]


In which case the mark scheme and you are also correct, you would attain full credit :smile:
Reply 8
Avatar for 01Chris02
01Chris02
OP
13
Original post by joostan
They're not both wrong.
recall that: −32ln⁡(2x)=−32ln⁡(2)+−32ln⁡(x)-\frac{3}{2}\ln(2x) = -\frac{3}{2}\ln(2) + -\frac{3}{2}\ln(x)
And also that:
=−32ln⁡(2) = -\frac{3}{2}\ln(2)
is just a constant.
You simply get a different constant (+c) after you integrate. :smile:


Thanks for clearing that up! :smile:
Reply 9
Avatar for joostan
joostan
13
Original post by 01Chris02
Thanks for clearing that up! :smile:


no problem :smile:

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