Question

(dfrac{left(4 + dfrac{1}{x + 1}
ight)}{left(16 - dfrac{1}{(x + 1)^2}
ight)})

Explanation:

Step1: Simplify numerator and denominator separately

First, simplify the numerator \(4+\frac{1}{x + 1}\). Find a common denominator, which is \(x + 1\). So we have \(\frac{4(x + 1)+1}{x + 1}=\frac{4x+4 + 1}{x + 1}=\frac{4x+5}{x + 1}\).

Next, simplify the denominator \(16-\frac{1}{(x + 1)^2}\). The common denominator is \((x + 1)^2\), so we get \(\frac{16(x + 1)^2-1}{(x + 1)^2}\). Notice that \(16(x + 1)^2-1\) is a difference of squares, where \(a = 4(x + 1)\) and \(b = 1\). So it factors as \([4(x + 1)-1][4(x + 1)+1]=(4x+4 - 1)(4x+4 + 1)=(4x + 3)(4x+5)\). Thus, the denominator becomes \(\frac{(4x + 3)(4x+5)}{(x + 1)^2}\).

Step2: Divide the simplified numerator by the simplified denominator

Now we have \(\frac{\frac{4x+5}{x + 1}}{\frac{(4x + 3)(4x+5)}{(x + 1)^2}}\). Dividing by a fraction is the same as multiplying by its reciprocal, so we get \(\frac{4x+5}{x + 1}\times\frac{(x + 1)^2}{(4x + 3)(4x+5)}\).

Step3: Cancel out common factors

We can cancel out the common factors \(4x + 5\) and one factor of \(x + 1\) from the numerator and denominator. After canceling, we are left with \(\frac{x + 1}{4x + 3}\), provided that \(4x+5
eq0\), \(x + 1
eq0\), and \(4x + 3
eq0\) (to avoid division by zero).

Answer:

\(\frac{x + 1}{4x + 3}\) (for \(x
eq - 1,-\frac{5}{4},-\frac{3}{4}\))