2. ## Re: Please help with these three questions they the only three I couldn't do from the

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

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3. ## Re: Please help with these three questions they the only three I couldn't do from the

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

4. ## Re: Please help with these three questions they the only three I couldn't do from the

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

5. ## Re: Please help with these three questions they the only three I couldn't do from the

I figured out 9d

6. ## Re: Please help with these three questions they the only three I couldn't do from the

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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7. ## Re: Please help with these three questions they the only three I couldn't do from the

C=0

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

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

8. ## Re: Please help with these three questions they the only three I couldn't do from the

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

9. ## Re: Please help with these three questions they the only three I couldn't do from the

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

10. ## Re: Please help with these three questions they the only three I couldn't do from the

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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11. ## Re: Please help with these three questions they the only three I couldn't do from the

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

12. ## Re: Please help with these three questions they the only three I couldn't do from the

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 $e$).

13. ## Re: Please help with these three questions they the only three I couldn't do from the

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
What does loge (e) equate to

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