What is expand 1/((1+x))-1/((2+x))+2/((2+x)^2) ?

The solution to expand 1/((1+x))-1/((2+x))+2/((2+x)^2) is (3x+4)/(x^3+5x^2+8x+4)

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expand 1/((1+x))-1/((2+x))+2/((2+x)^2)

expand

\frac{1}{( 1 + x )} - \frac{1}{( 2 + x )} + \frac{2}{( 2 + x ) ^{2}}

\frac{3 x + 4}{x ^{3} + 5 x ^{2} + 8 x + 4}

Solution steps

\frac{1}{1 + x} - \frac{1}{2 + x} + \frac{2}{( 2 + x ) ^{2}}

(2 + x)^{2} = 4 + 4 x + x^{2}

= \frac{1}{x + 1} - \frac{1}{x + 2} + \frac{2}{x ^{2} + 4 x + 4}

Factor

x^{2} + 4 x + 4 : (x + 2)^{2}

= \frac{1}{1 + x} - \frac{1}{2 + x} + \frac{2}{( x + 2 ) ^{2}}

Least Common Multiplier of

1 + x, 2 + x, (x + 2)^{2} : (x + 1) (x + 2)^{2}

Adjust Fractions based on the LCM

= \frac{( x + 2 ) ^{2}}{( x + 1 ) ( x + 2 ) ^{2}} - \frac{( x + 1 ) ( x + 2 )}{( x + 1 ) ( x + 2 ) ^{2}} + \frac{2 ( x + 1 )}{( x + 2 ) ^{2} ( x + 1 )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{( x + 2 ) ^{2} - ( x + 1 ) ( x + 2 ) + 2 ( x + 1 )}{( x + 1 ) ( x + 2 ) ^{2}}

(x + 2)^{2} = x^{2} + 4 x + 4

= \frac{( x + 2 ) ^{2} - ( x + 1 ) ( x + 2 ) + 2 ( x + 1 )}{( x + 1 ) ( x ^{2} + 4 x + 4 )}

Expand

(x + 1) (x^{2} + 4 x + 4) : x^{3} + 5 x^{2} + 8 x + 4

= \frac{( x + 2 ) ^{2} - ( x + 1 ) ( x + 2 ) + 2 ( x + 1 )}{x ^{3} + 5 x ^{2} + 8 x + 4}

Expand

(x + 2)^{2} - (x + 1) (x + 2) + 2 (x + 1) : 3 x + 4

= \frac{3 x + 4}{x ^{3} + 5 x ^{2} + 8 x + 4}

\frac{( 2 x + 3 ) ^{3}}{( 4 x ^{2} - 1 ) ^{8}}

\frac{100}{( s + 2 ) ( s + 3 ) ( s + 5 )}

(y - (- 4)) \cdot y^{2} (y = - 1)

sin (2 θ + θ)

\frac{6 x ^{2} + 4 x}{( x + 2 ) ^{2} ( x - 1 )}

What is expand 1/((1+x))-1/((2+x))+2/((2+x)^2) ?
The solution to expand 1/((1+x))-1/((2+x))+2/((2+x)^2) is (3x+4)/(x^3+5x^2+8x+4)