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C4 Integration Problem

For ∫12+2x\int \frac{1}{2+2x}, should we integrate it to 12\frac{1}{2}ln|1+x| or 12\frac{1}{2}ln|2+2x|??

Thank you in advance!!
Reply 1
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TeeEm
19
@Zacken
Reply 2
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Zacken
22
Original post by mystreet091234
For ∫12+2x\int \frac{1}{2+2x}, should we integrate it to 12\frac{1}{2}ln|1+x| or 12\frac{1}{2}ln|2+2x|??

Thank you in advance!!


The former:

∫dx2+2x=12∫dx1+x=12ln⁡∣1+x∣+C\displaystyle \int \frac{\mathrm{d}x}{2 + 2x} = \frac{1}{2} \int \frac{\mathrm{d}x}{1 + x} = \frac{1}{2 }\ln |1+ x| + \mathcal{C}
Reply 3
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ri916
1
The second one.
Original post by Zacken
The former:

∫dx2+2x=12∫dx1+x=12ln⁡∣1+x∣+C\displaystyle \int \frac{\mathrm{d}x}{2 + 2x} = \frac{1}{2} \int \frac{\mathrm{d}x}{1 + x} = \frac{1}{2 }\ln |1+ x| + \mathcal{C}


Where does dx come from??

THANKS!!
Reply 5
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Zacken
22
Original post by mystreet091234
Where does dx come from??

THANKS!!


You're integrating with respect to x.
Reply 6
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Ayman!
15
Original post by mystreet091234
For ∫12+2x\int \frac{1}{2+2x}, should we integrate it to 12\frac{1}{2}ln|1+x| or 12\frac{1}{2}ln|2+2x|??

Thank you in advance!!


Do note that
Unparseable latex formula:

\[\displaysyle \int \frac{f{(x)}'}{f\left ( x \right )}\mathrm{d}x = ln\left | f(x) \right | + C\]

.

Always try checking whether the denominator differentiates to the numerator. Taking out common factors might also help in certain cases.

Edit: it's equivalent, my apologies
(edited 8 years ago)
Original post by aymanzayedmannan
Do note that
Unparseable latex formula:

\[\displaysyle \int \frac{f{(x)}'}{f\left ( x \right )}\mathrm{d}x = ln\left | f(x) \right | + C\]

. Your second result does not satisfy this, which is why it's incorrect.

Always try checking whether the denominator differentiates to the numerator. Taking out common factors might also help in certain cases.


But when we let y=12\frac{1}{2}ln|2+2x|
dydx=12×12+2x×2=12+2x\frac{dy}{dx} = \frac{1}{2} \times \frac{1}{2+2x} \times2 = \frac{1}{2+2x}

So why is it wrong??
Original post by ri916
The second one.


Don't
Reply 9
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Zacken
22
Original post by mystreet091234
But when we let y=12\frac{1}{2}ln|2+2x|
dydx=12×12+2x×2=12+2x\frac{dy}{dx} = \frac{1}{2} \times \frac{1}{2+2x} \times2 = \frac{1}{2+2x}

So why is it wrong??


Ah, both answers are correct:

ln⁡∣2+2x∣+C=ln⁡2(1+x)+C=(ln⁡2+C)+ln⁡∣1+x∣=ln⁡∣1+x∣+C′\displaystyle \ln |2+2x| + \mathcal{C} = \ln 2(1+x) + \mathcal{C} = (\ln 2 + \mathcal{C}) + \ln |1+x| = \ln |1+x| + \mathcal{C}'

Where C′=C+ln⁡2\mathcal{C}' = \mathcal{C} + \ln 2, apologies.
Original post by Zacken
Ah, both answers are correct:

ln⁡∣2+2x∣+C=ln⁡2(1+x)+C=(ln⁡2+C)+ln⁡∣1+x∣=ln⁡∣1+x∣+C′\displaystyle \ln |2+2x| + \mathcal{C} = \ln 2(1+x) + \mathcal{C} = (\ln 2 + \mathcal{C}) + \ln |1+x| = \ln |1+x| + \mathcal{C}'

Where C′=C+ln⁡2\mathcal{C}' = \mathcal{C} + \ln 2, apologies.


Yeah we had this problem in class. You can have any constant factor in the log, right? As long as it gives a +ve result
Reply 11
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Zacken
22
Original post by Student403
Yeah we had this problem in class. You can have any constant factor in the log, right? As long as it gives a +ve result


Yep. :yep:
Million thanks to you all:smile:
(edited 8 years ago)

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