First find the length of the room by dividing the area by the width:
99 / 9 = 11
2 walls would be 9 feet and 2 walls would be 11 feet.
Perimeter is the sum of the 4 walls:
11 + 11 + 9 + 9 = 40 feet
Short answer: you don't.
The linear term in the numerator of the integral means the form shown is not applicable. Rather, you perform the integration using partial fraction expansion.
![\displaystyle\int{\frac{5x+1}{25x^2+60x-13}}\,dx=\int{\frac{5x+1}{(5x-1)(5x+13)}}\,dx\\\\=\frac{1}{35}\int{\frac{5}{5x-1}}\,dx+\frac{6}{35}\int{\frac{5}{5x+13}}\,dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%7B%5Cfrac%7B5x%2B1%7D%7B25x%5E2%2B60x-13%7D%7D%5C%2Cdx%3D%5Cint%7B%5Cfrac%7B5x%2B1%7D%7B%285x-1%29%285x%2B13%29%7D%7D%5C%2Cdx%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B35%7D%5Cint%7B%5Cfrac%7B5%7D%7B5x-1%7D%7D%5C%2Cdx%2B%5Cfrac%7B6%7D%7B35%7D%5Cint%7B%5Cfrac%7B5%7D%7B5x%2B13%7D%7D%5C%2Cdx)
The integral is ...
... (1/35)ln|5x-1| +(6/35)ln|5x+13| +C
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If the numerator of your integral were a constant, then the fractions multiplying the separate partial fraction integrals would have the same magnitude and opposite signs. You would end with the difference of logarithms, which could be expressed as the log of a ratio as shown in your problem statement.
Substitute n with 5 and 6
(3x5)+5 = 20
(3x6)+5 = 23
<span>F(3)=xx-3x+5
</span><span>F(3)=(3)(3)-3(3)+5
</span><span>F(3)=9-9+5
F(3)=5</span>
So first you need to combine like terms:
7b -2/5 = 6b - 7/5
+2/5 +2/5
7b = 6b -5/5
7b = 6b -1
-6b -6b
b=-1