# Please help with these three questions they the only three I couldn't do from the ex

• 14 Jan 2018, 6:29 PM
• 14 Jan 2018, 6:39 PM
ichila101
Quote:

You have already tried finding the indefinite integral for all three of them?

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• 14 Jan 2018, 6:42 PM
ichila101
Try differentiating the denominator of the first one and tell me whst you get

For the second one think of what you can do to a fraction when the numerator is made up of two different variables that are being added together

For the third one how can you turn the fraction into a reciprocal
• 14 Jan 2018, 6:50 PM
Is it y=ln (x^2+5x+4) + C

And then you substitute (1,1) to find C. But the solution is y=log (x^2+5x+4/10) +1
• 14 Jan 2018, 6:53 PM
I figured out 9d
• 14 Jan 2018, 6:59 PM
ichila101
Quote:

Is it y=ln (x^2+5x+4) + C

And then you substitute (1,1) to find C. But the solution is y=log (x^2+5x+4/10) +1

Yes thats right and what did you get when you substituted (1,1)

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• 14 Jan 2018, 7:03 PM
C=0

So y = log (x^2 +5x+4)

But the solution is as mentioned above y= log (x^2+5x+4/10) +1
• 14 Jan 2018, 7:07 PM
ichila101
Quote:

C=0

So y = log (x^2 +5x+4)

But the solution is as mentioned above y= log (x^2+5x+4/10) +1

You subbed the x=1 in log((1)^2 + 5(1) + 4)?
C shouldnt be equating to 0
• 14 Jan 2018, 7:21 PM
AAah okay I get what I did wrong thanks now I need 9e
• 14 Jan 2018, 7:26 PM
ichila101
Quote:

AAah okay I get what I did wrong thanks now I need 9e

Did you try turning it into reciprocal and then integrating it?

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• 14 Jan 2018, 7:32 PM
Okay I got 9e finally but how is y(e) = e+1!??

I got y= 2+x -log x

So y(e) = 2+e-log e
• 14 Jan 2018, 7:47 PM
fan96
Quote:

Okay I got 9e finally but how is y(e) = e+1!??

I got y= 2+x -log x

So y(e) = 2+e-log e

Evaluate $\log e$ (keeping in mind that logarithms are implicitly in base $e$).
• 14 Jan 2018, 7:50 PM
ichila101