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- #1

\(\displaystyle \int \frac{6z + 5}{2z + 1}dz\)

- Thread starter shamieh
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- Thread starter
- #1

\(\displaystyle \int \frac{6z + 5}{2z + 1}dz\)

- Nov 4, 2013

- 428

I don't see a reason for completing the square in this case. Its much easier to solve the problem at hand by rewriting it as:

\(\displaystyle \int \frac{6z + 5}{2z + 1}dz\)

$$\int 3+\frac{2}{2z+1} \,dz$$

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- #3

That's what I was thinking...But somehow they got \(\displaystyle \int \frac{3(2z + 1) +2}{2z + 1}\) and I'm just lost on how they get that in the numerator?I don't see a reason for completing the square in this case. Its much easier to solve the problem at hand by rewriting it as:

$$\int 3+\frac{2}{2z+1} \,dz$$

anyway you can show me so I can see ?

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Also, 6z/2z = 3 right? so wouldnt you have \(\displaystyle \int 3 + 5\) dx ??

- Nov 4, 2013

- 428

That's basically what I did. You can write $(a+b)/c=a/c+b/c$. So with a=3(2z+1), b=2 and c=2z+1,That's what I was thinking...But somehow they got \(\displaystyle \int \frac{3(2z + 1) +2}{2z + 1}\) and I'm just lost on how they get that in the numerator?

$$\int \frac{3(2z+1)+2}{2z+1}=\frac{3(2z+1)}{(2z+1)} + \frac {2} {(2z+1)}=3+\frac{2}{(2z+1)}$$

I hope that helps.

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- #5

- Nov 4, 2013

- 428

$$6z+5=6z+3+2$$The problem is I don't know how to get from \(\displaystyle 6z + 5\) to \(\displaystyle 3(2z + 1) + 2\)

Factor out 3 from the first two terms to get: $3(2z+1)+2$

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- #7

- Nov 4, 2013

- 428

Yes!So if I had for example, 6z + 7

I would say 6z + 7 = 6z + 4 + 3

and then get 6z + 7 = 2(3z + 2) + 3?

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- #9